a, \(|-2x|=3x+4\)

ĐK: 3x+4 \(\ge0\)

<=> 3x\(\ge-4\)

<=> x\(\ge\dfrac{-4}{3}\)

Ta có: \(\left[{}\begin{matrix}-2x=3x+4\\2x=3x+4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4}{5}\\x=-4\end{matrix}\right.\) mà x\(\ge\dfrac{-4}{3}\)

Vậy S=\(\left\{\dfrac{-4}{5}\right\}\)

b, \(|-2,5x|=5+1,5x\)

ĐK: 5+1,5x\(\ge0\)

<=> 1,5x\(\ge-5\)

<=> x\(\ge\dfrac{-10}{3}\)

Ta có: \(\left[{}\begin{matrix}-2,5x=5+1,5x\\2,5x=5+1,5x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1,25\\x=5\end{matrix}\right.\)

(thỏa mãn ĐK x\(\ge\dfrac{-10}{3}\))

Vậy S= \(\left\{-1,25;5\right\}\)

a,\(\left|-2x\right|=3x+4\)

\(Khi:-2x\ge0\Rightarrow x\le0\)

\(\Rightarrow-2x=3x+4\)

\(\Leftrightarrow-2x-3x=4\) \(\Leftrightarrow-5x=4\)

\(\Leftrightarrow x=\dfrac{-4}{5}\)(t/m)

Khi \(-2x< 0\Rightarrow x>0\)

\(\Rightarrow2x=3x+4\) \(\Leftrightarrow\) \(2x-3x=4\)

\(\Leftrightarrow-x=4\) \(\Leftrightarrow\) \(x=-4\)(k t/m)

Vậy tập nghiệm của pt là: \(S=\left\{-\dfrac{4}{5}\right\}\)

b. \(\left|-2,5x\right|=5+1,5x\)

Khi \(-2,5x\ge0\Rightarrow x\le0\)

\(\Rightarrow-2,5x=5+1,5x\)

\(\Leftrightarrow-2,5x-1,5x=5\) \(\Leftrightarrow\) \(-4x=5\Leftrightarrow x=-\dfrac{5}{4}\) (t/m)

Khi \(-2,5x< 0\Rightarrow x>0\)

\(\Rightarrow2,5x=5+1,5x\) \(\Leftrightarrow\) \(2,5x-1,5x=5\)

\(\Leftrightarrow x=5\)(t/m)

Vậy tập nghiệm của pt: \(S=\left\{-\dfrac{5}{4};5\right\}\)

1)\(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)

Vì 2≠0

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

Vậy: x∈{2;3}

3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

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

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

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

\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)

4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(=>11-3x+1=\frac{9}{2}-5+3,5x\)

\(=>-3x+12=3,5x-\frac{1}{2}\)

\(=>-3x-3,5x=-\frac{1}{2}-12\)

\(=>-6,5x=-12,5\)

\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)

Ủng hộ nha

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(11-3x+1=\frac{9}{2}-5+3,5x\)

\(12-3x=-\left(0,5\right)+3,5x\)

\(12,5-3x=3,5x\)

\(12,5=6,5x\)

\(x=12,5:6,5=\frac{25}{13}\)

Đề bài là gì vậy bạn?

phải cho ra kết quả một bên mới tính được x chứ

2.(x+1)-3x=2x-1

2x + 2-3x=2x-1

-x+2=2x-1

-x-2x=-1-2

-3x=-3

  x= 1

Vậy...

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

\(a,\)\(x^3-3x^2+1-3x\)

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

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

\(=\left(x+1\right)\left[\left(x+1\right)^2+3x\right]\)

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

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

\(b,\)\(3x-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(3x^2+3x\right)-\left(10x+10\right)\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(3x-10\right)\left(x+1\right)\)

\(c,\)\(x^4+1-2x^2\)

\(=x^4-x^2-x^2+1\)

\(=\left(x^4-x^2\right)-\left(x^2-1\right)\)

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

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

\(d,\)\(=x^2-3x+2\)

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

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

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

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