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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy....

=> ( x² -1 )( 2x +3) - ( x² - 1)( 3x +2 ) =0

=> (x² - 1). ( 2x +3 - 3x - 2) =0

=> ( x²- 1)( 1-x) = 0

=> x² - 1 =0 hoặc 1 - x =0

=> x= 1

Bài làm

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1\\x=-1\end{cases}}\)

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)

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

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

pt\(\Leftrightarrow\)x(\(\sqrt{x+1}-\sqrt{3x-1}\))+\(\sqrt{3x-1}\left(\sqrt{3x-1}-\sqrt{x+1}\right)\)=0

  \(\Leftrightarrow\)(\(\sqrt{x+1}-\sqrt{3x-1}\))(1-\(\sqrt{3x-1}\))=0

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

\(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)(Đk x≥\(\dfrac{1}{3}\))

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

=\(3x-1-x\sqrt{3x-1}+1\)

=\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

=\(\sqrt{x^2+2x+1-2+2x^2}-x\sqrt{x+1}+1\)

=\(\sqrt{\left(x+1\right)\left(3x-1\right)}-x\sqrt{x+1}+1\)

=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

ta có \(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)=\sqrt{x+1}\left(\sqrt{3x-1}-x\right)\)

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)

\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)

b, Sua de :  \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{1}{x}\)

Đề bài?

3

Lớp 3

Lời giải:

a. $3(x-\frac{1}{2})-3(x-\frac{1}{3})=x$

$3[(x-\frac{1}{2})-(x-\frac{1}{3})]=x$

$3.\frac{-1}{6}=x$

$\Rightarrow x=\frac{-1}{2}$

b.

$\frac{1}{2}(x+2)-4(x-\frac{1}{4})=\frac{1}{2}x$

$\frac{1}{2}x+1-4(x-\frac{1}{4})=\frac{1}{2}x$

$1-4(x-\frac{1}{4})=0$

$x-\frac{1}{4}=\frac{1}{4}$

$x=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Chiều cao hình thang : 15 . 1/3 = 5 (m)
Diện tích hình thang : 15.5:2 = 37,5(m^2)

Đ/s:...

(x)+Q(x)=(x³-2x+1)+(2x² -2x³+x-5)

 =x³-2x+1+2x²-2x³+x-5 = -x³+2x²-x-4

P(x)-Q(x)=(x³-2x+1)+(2x²-2x³+x-5) 

=x³-2x+1-2x²+2x³-x+5 

=3x³-2x²-3x+6

a)\(\frac{-1}{4}x^2y-\frac{1}{4}x^2y=-\frac{1}{2}x^2y.\)

thay x=1,y=-1 vào ta được:

\(-\frac{1}{2}.1^2.\left(-1\right)=\frac{1}{2}.\)

b)\(3x^2y^3+3x^2y^3=6x^2y^3.\)

thay x=1,y=-1 vào ta được:

\(6.1^2.\left(-1\right)^3=6.1.\left(-1\right)=-6.\)

c) \(6x^3y^4z-4x^3y^4z=2x^3y^4z.\)

Thay x=1,y=-1,z=2 vào ta được:

\(2.1^3.\left(-1\right)^4.2=2.1.1.2=4.\)

d) Thay x=1,y=-1,z=2 vào ta được:

\(1-2.\left(-1\right)^2+2^3=1-2+8=7.\)

Đầy đủ quá rồi đấy. Giữ lời hứa nha

Học tốt

1: =>x^2+4x+3-x^2-2x=7

=>2x=4

hay x=2