\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)

\(\Rightarrow2x-1=0\)hoặc \(2x-1=1\)hoặc \(2x-1=-1\)

\(\Rightarrow x=\frac{1}{2}\)hoặc \(x=1\)hoặc \(x=0\)

a)TH1: \(2x-3>0;3x+2>0\)

\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)

TH2: \(2x-3< 0;3x+2< 0\)

\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)

Cả 2 TH ra \(x=-5=>x=-5\)

b)TH1 \(\dfrac{1}{2}x>0\)

\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)

TH2 \(\dfrac{1}{2}x< 0\)

\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)

\(=>x=2;\dfrac{6}{5}\)

:V lập 2 ý là làm đc á em 

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

\(\Leftrightarrow2x\left[\left(x+2\right)^2-4x\right]=2\left(x^3-8\right)\)

\(\Leftrightarrow2x\left(x^2+4x+4-4x\right)=2x^3-16\)

\(\Leftrightarrow2x\left(x^2-4\right)=2x^3-16\)

\(\Leftrightarrow2x^3-8x=2x^3-16\)

\(\Leftrightarrow-8x=-16\)

\(\Leftrightarrow x=2\)

Mình k 3 k cho bạn rồi, nhưng bạn giải rõ ra cho mình được không? Đây là bài tập hè lớp 6 nhưng mình cảm giác giống bài lớp 8.

Bài làm:

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

\(\Leftrightarrow x^2-3x-2x^2+2x+x^2-5x=0\)

\(\Leftrightarrow-6x=0\)

\(\Rightarrow x=0\)

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

=> \(-x^2-x=-x^2+5x\)

=> \(-x^2+x^2=5x+x\)

=> 6x=0

=> x=0

Đúng hong tar? '-'

\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) \(123-5\left(x+4\right)=38\)

\(5\left(x+4\right)=123-38\)

\(5\left(x+4\right)=85\)

\(x+4=85:5\)

\(x+4=17\)

\(x=17-4\)

\(x=13\)

Vậy \(x=13\).

b) \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

\(3x-16=2\cdot7^4:7^3\)

\(3x-16=2\cdot7\)

\(3x-16=14\)

\(3x=14+16\)

\(3x=30\)

\(x=30:3\)

\(x=10\)

Vậy \(x=10\).

\(\left(2x-14\right)\left(3^x-9\right)=0\)

=> \(\begin{cases}2x-14=0\\3^x-9=0\end{cases}\) => \(\begin{cases}2x=14\\3^x=9\end{cases}\) => \(\begin{cases}x=7\\3^x=3^2\end{cases}\) => \(\begin{cases}x=7\\x=2\end{cases}\)

\(\left(2x-14\right)\left(3^x-9\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-14=0\\3^x-9=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=14\\3^x=9\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=7\\x=2\end{array}\right.\)

vừa đi đánh đàn 1 tí em đã làm của anh rùi :(

\(\left|x+123\right|+\left|\left|y+1\right|-5\right|=0\)

Ta có: \(\left|x+123\right|\ge0\) với mọi x

\(\left|\left|y+1\right|-5\right|\ge0\) với mọi y

Nên \(\left|x+123\right|+\left|\left|y+1\right|-5\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+123=0\\\left|y+1\right|-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-123\\\left|y+1\right|=5\end{matrix}\right.\)

\(\left|y+1\right|=5\Rightarrow y+1=\pm5\)

+ \(y+1=-5\Rightarrow y=-6\)

+ \(y+1=5\Rightarrow y=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-123\\\left[{}\begin{matrix}y=-6\\y=4\end{matrix}\right.\end{matrix}\right.\)

~ Học tốt ~