Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

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

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

a: \(\left|-2,5x\right|=x-12\)

=>\(2,5\left|x\right|=x-12\)

=>\(\begin{cases}x-12\ge0\\ \left(2,5x\right)^2=\left(x-12\right)^2\end{cases}\Rightarrow\begin{cases}x\ge12\\ \left(2,5x-x+12\right)\left(2,5x+x-12\right)=0\end{cases}\)

=>\(\begin{cases}x\ge12\\ \left(1,5x+12\right)\left(3,5x-12\right)=0\end{cases}\Rightarrow\begin{cases}x\ge12\\ x\in\left\lbrace-8;\frac{24}{7}\right\rbrace\end{cases}\)

=>x∈∅

b: |5x|-3x-2=0(1)

TH1: x>=0

=>5x>=0

(1) sẽ trở thành:

5x-3x-2=0

=>2x-2=0

=>2x=2

=>x=1(nhận)

TH2: x<0

=>5x<0

(1) sẽ trở thành:

-5x-3x-2=0

=>-8x=2

=>x=-1/4(nhận)

c: \(\left|-2x\right|+x-5x-3=0\)

=>2|x|-4x-3=0(2)

TH1: x>=0

(2) sẽ trở thành: 2x-4x-3=0

=>-2x-3=0

=>2x=-3

=>x=-3/2(loại)

TH2: x<0

(2) sẽ trở thành: -2x-4x-3=0

=>2x+4x+3=0

=>6x=-3

=>x=-1/2(nhận)

d: \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)

=>\(\left|x-3\right|+x^2-x^2-4x=0\)

=>|x-3|=4x

=>\(\begin{cases}4x\ge0\\ \left(4x\right)^2=\left(x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(4x-x+3\right)\left(4x+x-3\right)=0\end{cases}\)

=>\(\begin{cases}x\ge0\\ \left(3x+3\right)\left(5x-3\right)=0\end{cases}\)

=>x=3/5

e: \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)

=>\(x^2-2x+1+\left|x+21\right|-x^2-13=0\)

=>-2x-12+|x+21|=0

=>|x+21|=2x+12

=>\(\begin{cases}2x+12\ge0\\ \left(2x+12\right)^2=\left(x+21\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-6\\ \left(2x+12-x-21\right)\left(2x+12+x+21\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-6\\ \left(x-9\right)\left(3x+33\right)=0\end{cases}\Rightarrow x=9\)

lớp 7 có bài đó lun hẻ 

thế bn học lớp mấy

???

a/ \(5x^3-2x^2+4x-4\)

x \(x^3+3x^2-5x-1\)

  \(5x^6-6x^4-20x^2+4\)

b/ \(-4,2x^4+3,1x^2-\)\(\)7

 x       \(2,5x^3-\)7x + 1,5

\(-10,5x^{ }\)7\(-21,\)7\(x^3-10,5\)