cái này có vội không? nếu không thì sáng mai mình giải cho bạn?

1)\(\left(x-5\right)^2+\left(x+3\right)^2=2\cdot\left(x-4\right)\cdot\left(x+4\right)-5x+7\)

\(\Leftrightarrow x^2-10x+25+x^2+6x+9-2x^2+32+5x-7=0\)

\(\Leftrightarrow x+59=0\)

\(\Rightarrow x=-59\)

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)

\(\Rightarrow2x-5=0\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=\dfrac{5}{2}\)

\(b,2x^3+3x^2+2x+3=0\)

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

\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

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

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

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

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

\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)

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

\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)

a)\(\frac{3}{x-4}-\frac{2}{4-x}=\frac{3}{x-4}+\frac{2}{x-4}=\frac{5}{x-4}\)

câu b làm tương tự nha bạn

c)\(\frac{3}{x+5}-\frac{2}{x+2}=\frac{3x+6-2x-10}{\left(x+5\right)\left(x+2\right)}=\frac{x-4}{\left(x+5\right)\left(x+2\right)}\)

d)\(\frac{9}{x-5}-\frac{6}{x^2-25}=\frac{9x+45-6}{x^2-25}=\frac{9x+39}{x^2-25}\)

mik làm hơi tắt bạn thông cảm nha

a)

\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)

Vậy \(x=\left\{\frac{7}{19}\right\}\) là nghiệm của phương trình.

b)

\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3x-6}{x-2}-\frac{3-x}{x-2}=0\\ \Leftrightarrow1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Rightarrow x=\frac{8}{4}=2\)

Mà \(ĐKXĐ:x\ne2\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

c)

\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8x-56}{x-7}-\frac{1}{x-7}=0\\ \Leftrightarrow8-x-8x+56-1=0\\ \Leftrightarrow63-9x=0\\ \Rightarrow x=\frac{63}{9}=7\)

Mà \(ĐKXĐ:x\ne7\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

d)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\\ \Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2-20=0\\ \Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Rightarrow x=1\)

Vậy \(x=1\) là nghiệm của phương trình.

e)

\(\frac{x}{2\cdot\left(x-3\right)}+\frac{x}{2\cdot\left(x+1\right)}=\frac{2x}{\left(x+1\right)\cdot\left(x-3\right)}\\ \Leftrightarrow\frac{x^2+x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}+\frac{x^2-3x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}-\frac{4x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Mà \(ĐKXĐ:x\ne\left\{-1;3\right\}\\ \Rightarrow x=0\)

Vật \(x=0\) là nghiệm của phương trình.

f)

\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\\ \Leftrightarrow\frac{12x+120}{12x\cdot\left(x+10\right)}+\frac{12x}{12x\cdot\left(x+10\right)}-\frac{x^2+10x}{12x\cdot\left(x+10\right)}=0\\ \Leftrightarrow12x+120+12x-x^2-10x=0\\ \Leftrightarrow14x+120-x^2=0\\ \Leftrightarrow x^2-14x-120=0\\ \Leftrightarrow x^2+6x-20x-120=0\\ \Leftrightarrow x\cdot\left(x+6\right)-20\cdot\left(x+6\right)=0\\ \Leftrightarrow\left(x-20\right)\cdot\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-20=0\\x+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

Vậy \(x=\left\{20;-6\right\}\) là ngiệm của phương trình.

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

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

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

a) \(\dfrac{3}{x-4}-\dfrac{2}{4-x}\)

\(=\dfrac{3}{x-4}+\dfrac{2}{x-4}\)

\(=\dfrac{3+2}{x-4}\)

\(=\dfrac{5}{x-4}\)

b) \(\dfrac{7}{x-3}-\dfrac{4}{3-x}\)

\(=\dfrac{7}{x-3}+\dfrac{4}{x-3}\)

\(=\dfrac{7+4}{x-3}\)

\(=\dfrac{11}{x-3}\)

c) \(\dfrac{3}{x-5}-\dfrac{2}{x+2}\) MTC: \(\left(x-5\right)\left(x+2\right)\)

\(=\dfrac{3\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{3\left(x+2\right)-2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{3x+6-2x+10}{\left(x-5\right)\left(x+2\right)}\)

\(=\dfrac{x+16}{\left(x-5\right)\left(x+2\right)}\)

d) \(\dfrac{9}{x-5}-\dfrac{6}{x^2-25}\)

\(=\dfrac{9}{x-5}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\) MTC: \(\left(x-5\right)\left(x+5\right)\)

\(=\dfrac{9\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{9\left(x+5\right)-6}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{9x+45-6}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{9x+39}{\left(x-5\right)\left(x+5\right)}\)

a) \(25x^2-9=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{3}{5}\end{cases}}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

\(\Leftrightarrow x=-\frac{1}{8}\)

a) ko hiểu đề bài

b) Ta có (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 8x + 16 - (x² - 1) = 16

<=>  x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

=> 8x = -1

=> x = \(-\frac{1}{8}\)

a) Ta có:

1/x+1/2x=3/2

2/2x+1/2x=3/2

3/2x=3/2

=>2x=2

=>x=1

Vậy x=1

#Học tốt