nhân chéo là đc:

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x     =14

 x      =2

Vậy x=2

a/ ĐKXĐ: ...

\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)

TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)

TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)

\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)

\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)

Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)

\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)

\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)

Vế trái luôn dương nên pt vô nghiệm

b/ ĐKXĐ: ...

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

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)

\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)

Thế vào pt dưới:

\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)

Đặt \(x+\frac{1}{x}+1=t\)

\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)

\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)

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

\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)

\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)

\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)

\(\Leftrightarrow2\sqrt{x-8}+16=x\)

\(\Leftrightarrow x=24\)

đơn giản 

nhưng trả lời câu hỏi của tớ đã

\(\frac{x+3}{-4}=-\frac{9}{x+3}\)

\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x+3\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)

Vậy ....

quy đồng

\(\left(x+3\right)^2=36\)

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

áp dụng định lí "  \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được

\(\left(x+3-6\right)\left(x+3+6\right)=0\)

\(x=3,x=-9\)

nhân chéo: 

(x+3)(x+3)=-4.(-9)

(x+3)^2   =36

Hai TH: 

TH1: x+3=6

=>   x=3

TH2: x+3=-6

=>  x=-9

Vậy x=3 hoặc x=-9

b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)

\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)

\(2,4.x-36=-\frac{12}{7}\)

\(2,4.x=-\frac{12}{7}+36\)

\(2,4.x=\frac{240}{7}\)

\(x=\frac{240}{7}\div2,4\)

\(x=\frac{100}{7}\)

a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

Câu c tương tự nhé. 

1) Tìm x

\(\frac{11}{2}.x+\frac{1}{3}.x=1\)

\(\Rightarrow x\left(\frac{11}{2}+\frac{1}{3}\right)=1\)

\(\Rightarrow x\left(\frac{33}{6}+\frac{2}{6}\right)=1\)

\(\Rightarrow x.\frac{35}{6}=1\)

\(\Rightarrow x=\frac{6}{35}\)

2) So sánh

\(\frac{59}{40}< \frac{50}{31}\)( cái này bạn quy đồng là ra, mik chỉ ghi kq, bạn tự tính )

3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)

\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}\right)-\frac{1}{2}\)

\(=-\frac{1}{3}+\frac{3}{14}-\frac{1}{2}\)

\(=-\frac{13}{21}\)

1)\(\frac{11}{2}.x+\frac{1}{3}.x=1\)

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

\(x.\frac{35}{6}=1\)

\(x=1:\frac{35}{6}\)

\(x=\frac{6}{35}\)

2) Ta có:

\(\frac{59}{40}=\frac{1829}{1240}\)

\(\frac{50}{31}=\frac{2000}{1240}\)

Vì \(2000>1829\Rightarrow\frac{2000}{1240}>\frac{1829}{1240}\Rightarrow\frac{50}{31}>\frac{59}{40}\)

3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)

\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}-\frac{1}{2}\right)\)

\(=-\frac{1}{3}+\left(\frac{8}{14}-\frac{5}{14}-\frac{7}{14}\right)\)

\(=\frac{-1}{3}+\frac{-4}{14}\)

\(=\frac{-1}{3}+\frac{-2}{7}\)

\(=\frac{-7}{21}+\frac{-6}{21}\)

\(=\frac{-13}{21}\)

A= 2x^2 + 4x + xy + 2y 

=(xy+2x²)+(2y+4x)

=x(y+2x)+2(y+2x)

=(x+2)(y+2x)

Thay x=88,y=-76 ta được:

A=(88+2)*(-76+2*88)

=90*100

=9 000

B= x^2 +xy - 7x - 7y

=(xy-7y)+(x²-7x)

=y(x-7)+x(x-7)

=(x-7)(y+x).Thay vào tính bình thường 

Cho mình làm phiền 1 xíu nữa thôi câu b vừa giải cho mình có mấy dấu * nghĩa là gì vậy