a) (x + 5)² - (x - 3)² = 2x - 7

(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7

8(2x + 2)= 2x - 7

16x + 16 = 2x - 7

16x - 2x = - 7 - 16

14x = - 23

x = - 23/14

b) (2x - 3)(4x² + 6x + 9) = 98

(2x)³ - 3³ = 98

8x³ - 27 = 98

8x³ = 125

x³ = 125/8

x³ = (5/2)³

x = 5/2

bài 1:

b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)

<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)

=>\(x^2+4x+4=x^2+5x+4+x^2\)

<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)

<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)

vậy...............

d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

vậy............

bài 3:

g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

=>\(4x-8-2x-2=x+3\)

<=>\(x=13\)

vậy..............

mấy ý khác bạn làm tương tụ nhé

chúc bạn học tốt ^ ^

1) \(\left(x-3\right)\left(x-5\right)+44\)

\(=x^2-3x-5x+15+44\)

\(=x^2-8x+59\)

\(=x^2-2.x.4+4^2+43\)

\(=\left(x-4\right)^2+43\ge43>0\)

\(\rightarrowĐPCM.\)

2) \(x^2+y^2-8x+4y+31\)

\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)

\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)

\(\rightarrowĐPCM.\)

3)\(16x^2+6x+25\)

\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)

\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)

\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)

\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)

-> ĐPCM.

4) Tương tự câu 3)

5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)

\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)

\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)

-> ĐPCM.

6) Tương tự câu 5)

7) 8) 9) Tương tự câu 3).

Giải rõ giúp mình với

\(a,x^4-16x^2+32x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)

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

\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)

câu nào dễ xơi trước

g) \(x^3+3x^2-2x-6=0\Leftrightarrow x^2\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\x=-3\end{matrix}\right.\)

kl: ...........

Mình chỉ biết bài b) thôi, mà cũng ko biết có đúng ko

x⁴+x³+x+1=0

<=> (x⁴+x³)+(x+1)=0

<=> x³(x+1)+(x+1)

<=> (x+1)(x³+1)=0

=>x+1=0

    x³+1=0

=> x= -1

     x³= -1

=> x= -1

a) \(x^2+6x+9=144\)

\(\Leftrightarrow\left(x+3\right)^2=12^2\)

\(\Leftrightarrow x+3=12\)

\(\Leftrightarrow x=9\)

\(\text{a) }x^2+6x+9=144\\ \Leftrightarrow\left(x^2+6x+9\right)-144=0\\ \Leftrightarrow\left(x+3\right)^2-12^2=0\\ \Leftrightarrow\left(x+3+12\right)\left(x+3-12\right)=0\\ \Leftrightarrow\left(x+15\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+15=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=9\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{9;-15\right\}\)

\(\dfrac{x-19}{1999}+\dfrac{x-23}{1995}+\dfrac{x+82}{700}=5\\ \Leftrightarrow\left(\dfrac{x-19}{1999}-1\right)+\left(\dfrac{x-23}{1995}-1\right)+\left(\dfrac{x+82}{700}-3\right)=0\\ \Leftrightarrow\dfrac{x-2018}{1999}+\dfrac{x-2018}{1995}+\dfrac{x-2018}{700}=0\\ \Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\right)=0\\ \Leftrightarrow x-2018=0\left(\text{Vì }\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\ne0\right)\\ \Leftrightarrow x=2018\)

Vậy nghiệm của phương trình là \(x=2018\)

\(\text{c) }x^3-3x^2+4=0\\ \Leftrightarrow x^3-2x^2-x^2+4=0\\ \Leftrightarrow\left(x^3-2x^2\right)-\left(x^2-4\right)=0\\ \Leftrightarrow x^2\left(x-2\right)-\left(x+2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-2x+x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-2x\right)+\left(x-2\right)\right]\left(x-2\right)=0\\ \Leftrightarrow\left[x\left(x-2\right)+\left(x-2\right)\right]\left(x-2\right)=0\\\Leftrightarrow \left(x+2\right)\left(x-2\right)^2=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right. \)

Vậy tập nghiệm phương trình là \(S=\left\{-2;2\right\}\)

a) (2x-4)(x²-16)=0

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\pm4\end{cases}}}\)

Vậy..

b) (x+5)²-25=0

\(\left(x+5\right)^2=25\)

\(\left(x+5\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x+5=5\\x+5=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}}\)

Vậy..

c) x²-6x+9=0

\(x.\left(1-6\right)=-9\)

\(x.\left(-5\right)=-9\)

\(x=\frac{9}{5}\)

chúc bạn học tốt !!!!

Đặt x làm thừa số chung là ra đó bạn

phương trình đa thức đối xứng

a) x² + 10x + 25 - 4x² - 20x = 0

<=> 3x² + 10x - 25 = 0

<=> (x + 5)(3x - 5) = 0 <=> \(\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)

Vậy S = \(\left\{-5;\frac{5}{3}\right\}\)

b. (4x - 5)² - 2(4x - 5)(4x + 5) = 0

<=> (4x - 5)[(4x - 5) - 2(4x + 5)] = 0

<=> (4x - 5)(4x - 5 - 8x - 10) = 0

<=> (4x - 5)(-4x - 15) = 0 <=> \(\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}\)

Vậy S = \(\left\{-\frac{15}{4};\frac{5}{4}\right\}\)