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1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
1) \(2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+x-2\)
\(=\left(2x+1\right)\left(x-2\right)\)
2) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(3x-10\right)\left(x+1\right)\)
1. <=> \(\left(3x+2\right)^3-\left(\left(3x\right)^3+2^3\right)=0\)
<=> \(\left(\left(3x\right)^3+2^3+3\left(3x+2\right).3x.2\right)-\left(\left(3x\right)^3+2^3\right)=0\)
<=>3 (3x + 2) . 3x.2 = 0
<=> (3x + 2 ) . x = 0
<=> x = -2/3 hoặc x = 0
2. Tương tự
1
\(\left(3x+2\right)^3-\left[\left(3x\right)^3+2^3\right]=0\)
\(\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2+2^3-\left(3x\right)^3-2^3=0\)
\(54x^2+36x=0\)
\(18x\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-2}{3}\end{cases}}\)
2
\(\left(2x+1\right)^3-\left[\left(2x\right)^3-1^3\right]=0\)
\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3-\left(2x\right)^3-1^3=0\)
\(12x^2+6x=0\)
\(6x\left(2x+1\right)=0\)
\(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
đề là gì
a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)
vậy x=2/3 hoặc x=-6
a, (3x-2) (x+6) (x^2 +5) = 0
<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x2 + 5 = 0 (loại vì x2 \(\ge\)0 => x2 + 5 > 0)
<=> x = 2/3 hoặc x = -6
b, (2x+5)^2 = (3x-1)^2
<=> (2x + 5)2 - (3x - 1)2 = 0
<=> (2x + 5 - 3x + 1)(2x + 5 + 3x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)
c, 4x2 (x-1) - x+1 = 0
<=> 4x2(x - 1) - (x - 1) = 0
<=> (x - 1)(4x2 - 1) = 0
<=> (x - 1)(2x - 1)(2x + 1) = 0
vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0
hay x = 1 hoặc x = 1/2 hoặc x = -1/2
a)x2+(x-3)(3x-5)=9
<=>x2+3x2-5x-9x+15=9
,<=>4x2-14x+15=9
<=>4x2-14x+6=0
<=>4x2-12x-2x+6=0
<=>4x(x-3)-2(x-3)=0
<=>(x-3)(4x-2)=0
=> x-3=0 hoặc 4x-2=0 =>x=3 hoặc x=1/2
b)(3x+2)2=(x-4)2
<=>(3x+2)2-(x-4)2=0
<=>(3x+2-x+4)(3x+2+x-4)=0 (HẰNG ĐẲNG THỨC SỐ 3)
<=>(2x+6)(4x-2)=0
=>2x+6=0 hoặc 4x-2 => x=-3 hoặc x=1/2
c)Chưa ra thông cảm ahihi
c, x4+2x3-2x2+2x-3 = 0
<=> (x4-x3)+(3x3-3x2)+(x2-x)+(3x-3) = 0
<=> x3(x-1)+3x2(x-1)+x(x-1)+3(x-1) = 0
<=> (x-1)(x3+3x2+x+3) = 0
<=> (x-1)[x2(x+3)+(x+3)] = 0
<=> (x-1)(x+3)(x2+1) = 0
<=> x-1 =0 hoặc x+3=0 ( vì x2+1 khác 0 )
<=> x =1 hoặc x= -3
a) \(x^3+x^2+2x-16\ge0\)
\(\Leftrightarrow x^3-2x^2+3x^2-6x+8x-16\ge0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+8\right)\ge0\)
Mà \(x^2+3x+8>x^2+3x+2,25=\left(x+1,5\right)^2\ge0\)
Cho nên \(x-2\ge0\)
\(\Leftrightarrow x\ge2\)
a,x^3-2x^2+3x^2-6x+8x-16>=0
(x^2+3x+8)(x-2)>=0
x^2+3x+8>0
=> để lớn hơn hoac bang 0 thì x-2 phải>=0
=>x>=2
b,hình như là vô nghiệm ko chắc chắn lắm
123Kendjdjdjdk
Ta có: \(2x^3-3x=0\)
=>\(x\left(2x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x^2-3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{3}{2}\end{matrix}\right.\)
=>\(x\in\left\{0;\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)