Mới lớp 8, chịu

Mà hình như trong pt phân số thứ 2 thiếu bình phương thì phải

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

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

\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

d, 2x²-5x-3 = 0

\(\Leftrightarrow\)2x²-6x+x-3= 0

\(\Leftrightarrow\)(2x²-6x) +(x-3) = 0

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

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

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

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

Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)

a) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\) Đk : x \(\ne-1\) ; x \(\ne2\)

\(\Leftrightarrow\frac{2.\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}-\frac{1.\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow2x-3x-x=-11+4+1\)

\(\Leftrightarrow-2x=-6\)

\(\Leftrightarrow x=3\)

Vậy S = \(\left\{3\right\}\)

Ta có : \(\frac{2}{2x-6}+\frac{1}{x+2}+\frac{2.x}{\left(x+1\right).\left(3-x\right)}=0\)

ĐKXĐ : x \(\ne\)-1 ; x \(\ne\)-2 ; x \(\ne\)3

MTC : ( x + 1 ) . ( x+ 2 ) . ( x - 3 ) 

<=> ( x + 1 ) . ( x + 2 ) + ( x + 1 ) . ( x + 3 ) - 2.x. ( x + 2 ) = 0

<=> x² + x + 2.x + 2 + x² -3.x + x -3 - 2.x² -4.x               = 0

<=> -3.x                                                                             = 1

<=> x = \(\frac{-1}{3}\)

Vậy S = { ​​​\(\frac{-1}{3}\)​}

ĐKXĐ: x khác 3, x khác -1

\(\frac{2}{2x-6}+\frac{2}{2x+2}+\frac{2}{\left(x+1\right)\left(3-x\right)}=0\)

<=> \(\frac{-1}{3-x}+\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(3-x\right)}=0\)

<=> \(\frac{-x-1}{\left(3-x\right)\left(x+1\right)}+\frac{3-x}{\left(3-1\right)\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(3-x\right)}=0\)

<=> \(\frac{-2x+4}{\left(3-x\right)\left(x+1\right)}=0\)

<=> -2x+4=0

<=>x=-2

vậy ....