làm nốt

d) (2x-1)(3x+2)(3-x)

=(6x²+x-2)(3-x)

=-6x³+17x²+5x-6

e) (x+3)(x²+3x-5)

=x³+6x²+4x-15

f) (xy-2)(x³-2x-6)

=x⁴y-2x³-2x²y-6xy+4x+12

g) (5x³-x2+2x-3)(4x²-x+2)

=20x⁵-9x⁴+19x³-16x²+7x-6

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x²+4x-10x-8

=5x²-6x-8

Bài 1:

1 (x+3)²=x²+6x+9

2

a, 2x²(3x-5x³)+10x⁵-5x³=6x³-10x⁵+10x⁵-5x³=x³

b, (x+3)(x²-3x+9)+(x-9)(x+3)=(x³+27)+(x²-6x-27)=x³+x²-6x

Bài 2:

a, x²-25x=0

\(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\begin{cases}x=0\\x-25=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=0\\x=25\end{cases}\)

b, (4x-1)²-9=0

\(\Leftrightarrow\left(4x-1-3\right)\left(4x-1+3\right)=0\)

\(\Leftrightarrow\left(4x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow4\left(x-1\right)2\left(2x+1\right)=0\)

\(\Leftrightarrow8\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\begin{cases}x-1=0\\2x+1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}\)

Bài 3:

a, 3x²-18x+27=3(x²-6x+9)=3(x-3)²

b, xy-y²-x+y=y(x-y)-(x-y)=(y-1)(x-y)

c, x²-5x-6=x²-6x+x-6=x(x-6)+(x-6)=(x+1)(x-6)

Bài 4:

a, ( 12x³y³-3x²y³+4x²y⁴):6x²y³=(12x³y³:6x²y³)-(3x²y³:6x²y³)+(4x²y⁴:6x²y³)

=2x-1/2 + 2/3y

b, bạn ơi mình không biết cách vẽ đường kẻ để chia ý , nếu bạn biết thì chỉ cho mình rồi mình làm cho

Bài 5 :

b, A = x(2x-3)

A= 2x²-3x

A= 2(x²-3/2x)

A= 2(x²-2x3/4+9/16-9/16)

A=2[(x-3/4)²-9/16]

A=2(x-3/4)²-9/8

A=2(x-3/4)²+(-9/8)

Vì (x-3/4)² \(\ge\)0 \(\forall x\)

-> 2(x-3/4)² \(\ge0\forall x\)

-> 2(x-3/4)²+(-9/8)\(\ge-\frac{9}{8}\forall x\)

Vậy MinA= -9/8

Bài 1:

1. Khai triển hằng đẳng thức

(x+3)² = x²+6x+9

2. Thực hiện phép tính

a) 2x²(3x-5x³)+10x⁵-5x³

=6x³-10x⁵+10x⁵-5x³

=x³

b)(x+3)(x²-3x+9)+(x-9)(x+3)

=(x³+27)+(x²+3x-9x-27)

=x³+27+x²+3x-9x-27

=x³+x²-6x

Bài 2:

a) x²-25x=0

\(\Leftrightarrow\)x(x-25)=0

\(\Leftrightarrow\) \(\left[\begin{matrix}x=0\\x-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=0\\x=25\end{matrix}\right.\)

Vậy x=0 hoặc x=25

b)(4x-1)² - 9=0

\(\Leftrightarrow\)(4x-1+3)(4x-1-3)=0

\(\Leftrightarrow\)(4x+2)(4x-4)=0

\(\Leftrightarrow\)2(2x+1)(2x-2)=0

\(\Leftrightarrow\left[\begin{matrix}2x+1=0\\2x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=\frac{-1}{2}\\x=1\end{matrix}\right.\)

Vậy x=1 hoặc x=\(\frac{-1}{2}\)

Bài 3:

a) 3x²-18x+27

=3(x²-6x+9)

=3(x-3)²

b) xy-y²-x+y

=(xy-y²)-(x-y)

=y(x-y)-(x-y)

=(x-y)(y-1)

c) x²-5x-6

=x²-6x+x-6

=(x²-6x)+(x-6)

=x(x-6)+(x-6

=(x-6)(x+1)

Bài 4:

a) (12x³y³-3x²y³+4x²y⁴) : 6x²y³

=x²y³(12x-3+4y): 6x²y³

=(12x-3+4y) : 6

= (12x : 6)-(3 : 6)+(4y : 6)

=2x-\(\frac{1}{2}\)+\(\frac{2y}{3}\)

b) (6x³-19x²+23x-12) : (2x-3)

=(3x²-5x+4)(2x-3) : (2x-3)

=3x²-5x+4

a) (2x−1).(2x+1)−4x^2=3

     4x²+2x-2x-1-4x²=3

     -1=3(bn xem lại đề đi)

Không phải xem bạn ạ, làm đến bước đó mở ngoặc ghi vô lí rôi kết luận ko có gt của x

a)(2x-1)(2x+1)-4x²=3

\(\Rightarrow\)4x²-1-4x²=3

\(\Rightarrow\)4x²-4x²=3+1

\(\Rightarrow\)0x=3

\(\Rightarrow\)x=0

Bài 1:

a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)

\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)

\(\Leftrightarrow-5=0\)(vl)

Vậy: \(x\in\varnothing\)

b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)

hay x=1

Vậy: x=1

c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)

\(\Leftrightarrow2x-72=0\)

\(\Leftrightarrow2\left(x-36\right)=0\)

mà 2>0

nên x-36=0

hay x=36

Vậy: x=36

d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)

\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)

\(\Leftrightarrow120x+36=56-64x\)

\(\Leftrightarrow120x+36-56+64x=0\)

\(\Leftrightarrow184x-20=0\)

\(\Leftrightarrow184x=20\)

hay \(x=\frac{5}{46}\)

Vậy: \(x=\frac{5}{46}\)

e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)

\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)

\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)

\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)

\(\Leftrightarrow-23x+29=0\)

\(\Leftrightarrow-23x=-29\)

hay \(x=\frac{29}{23}\)

Vậy: \(x=\frac{29}{23}\)

f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)

\(\Leftrightarrow2x+8-10x-50-25=0\)

\(\Leftrightarrow-8x-67=0\)

\(\Leftrightarrow-8x=67\)

hay \(x=\frac{-67}{8}\)

Vậy: \(x=\frac{-67}{8}\)

g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)

\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)

\(\Leftrightarrow10-5x-8x-8+12x-30=0\)

\(\Leftrightarrow-x-28=0\)

\(\Leftrightarrow-x=28\)

hay x=-28

Vậy: x=-28

h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)

\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)

\(\Leftrightarrow0x=0\)

Vậy: \(x\in R\)

Bài 2:

a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)

b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)

c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)

\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: Tập nghiệm S={-3}

d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)

\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)

\(\Leftrightarrow12-7x=0\)

\(\Leftrightarrow7x=12\)

hay \(x=\frac{12}{7}\)

Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)

e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x

\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow31x=1\)

hay \(x=\frac{1}{31}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)

Bài 2:

a)\(x^3-2x^2+x\)

\(=x\left(x^2-2x+1\right)\)

\(=x\left(x-1\right)^2\)

b)\(x^2-2x-15\)

\(=x^2-5x+3x-15\)

\(=x\left(x-5\right)+3\left(x-5\right)\)

c)\(y\left(x-z\right)+7\left(z-x\right)\)

\(=7\left(z-x\right)-y\left(z-x\right)\)

\(=\left(7-y\right)\left(z-x\right)\)

\(=\left(x-5\right)\left(x+3\right)\)

d)\(36-12x+x^2\)

\(=x^2-12x+36\)

\(=\left(x-6\right)^2\)

Bài 1:

a)\(2x\left(x^2-7x-3\right)=2x^3-14x^2-6x\)

b)\(\left(-2x^3+34y^2-7xy\right)\cdot4xy^2=136xy^4-28x^2y^3-8x^4y^2\)

c)\(\left(x^2-2x+3\right)\left(x-4\right)\)

\(=x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)\)

\(=x^3-4x^2-2x^2+8x+3x-12\)

\(=x^3-6x^2+11x-12\)

d)\(\left(2x^3-3x-1\right)\left(5x+2\right)\)

\(=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)\)

\(=10x^4-15x^2-5x+4x^3-6x-2\)

\(=10x^4+4x^3-15x^2-11x-2\)