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10 tháng 1 2020

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

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

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

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

\(\Rightarrow x^2+x-2=x^2-2x-2x+4\)

\(\Rightarrow x^2+x-x^2+2x+2x=4+2\)

\(\Rightarrow5x=6\)

\(\Rightarrow x=\frac{6}{5}\)

17 tháng 8 2018

(2x - 3)3 - 59 = 5

(2x - 3)3 = 5 + 59

(2x - 3)3 = 64

(2x - 3)3 = 43

=> 2x - 3 = 4

=> 2x = 4 + 3

=> 2x = 7

=> x ko có giá trị

5 tháng 8 2020

Bài 9 : Tìm x, biết :

a, (x - 2)(x - 3) + (x - 2) - 1 = 0

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

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy x ={1; 3}

b, (x + 2)2 - 2x(2x + 3) = (x + 1)2

\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)

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

\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x

\(\Leftrightarrow6x^3+x^2-2x=0\)

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

\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)

25 tháng 7 2021

a) (x-2)3+6(x+1)2-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

25 tháng 7 2021

b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2

\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2

\(\Rightarrow\)x2​-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2

\(\Rightarrow\)3x2-18x-22=3x2+2x+17

\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

16 tháng 10 2022

Câu 1: 

a: TH1: x>=6

A=12x-5-x+6=11x+1

TH2: x<6

A=12x-5+x-6=13x-11

b: TH1: x>=6

=>11x+1=11

=>11x=10

=>x=10/11(loại)

TH2: x<6

=>13x-11=11

=>13x=22

=>x=22/13(nhận)