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\(\frac{64x^3+1}{16x^2-1}=\frac{A}{4x-1}\left(x\ne\pm\frac{1}{4}\right)\)
\(\Leftrightarrow\frac{\left(4x+1\right)\left(16x^2+4x+1\right)}{\left(4x+1\right)\left(4x-1\right)}=\frac{A}{4x-1}\)
\(\Leftrightarrow\frac{\left(16x^2+4x+1\right)}{\left(4x-1\right)}=\frac{A}{4x-1}\)
Vậy \(A=\left(16x^2+4x+1\right)\)
\(\frac{4x^2+3x-7}{B}=\frac{4x+7}{2x-3}\left(x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{4x^2+7x-4x-7}{B}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{x\left(4x+7\right)-\left(4x+7\right)}{B}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(4x+7\right)}{B}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{\left(x-1\right)}{B}=\frac{1}{2x-3}\)
\(\Leftrightarrow B=\left(x-1\right)\left(2x-3\right)=2x^2-5x+3\)
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
ĐKXĐ bạn tự tìm nha : )
k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)
\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)
i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)
\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)
h, = k,
f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
a)\(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(84x+63-90x+30=175x+140+315\)
93-6x=175x+455
-362=181x
x=-2
b)\(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\cdot\left(2x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)
\(=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)
c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)
\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)
d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)
\(=\frac{3-12x^2}{-2x^2-4x+16}\)
a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)
\(=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}\)
Bài 4:
a) \(\frac{2x^2-10xy}{2xy}+\frac{5y-x}{y}\)
\(=\frac{y.\left(2x^2-10xy\right)}{2xy.y}+\frac{2xy.\left(5y-x\right)}{2xy.y}\)
\(=\frac{2x^2y-10xy^2}{2xy^2}+\frac{10xy^2-2x^2y}{2xy^2}\)
\(=\frac{2x^2y-10xy^2+10xy^2-2x^2y}{2xy^2}\)
\(=\frac{0}{2xy^2}\)
\(=0.\)
b) \(\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{x^2-y^2}\)
\(=\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{1.\left(x+y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y}{\left(x-y\right).\left(x+y\right)}+\frac{x+y}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y+x+y+3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x-y}{\left(x-y\right).\left(x+y\right)}\)
c) \(x+y+\frac{x^2+y^2}{x+y}\)
\(=\frac{x+y}{1}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right).\left(x+y\right)}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right)^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2+x^2+y^2}{x+y}\)
\(=\frac{2x^2+2xy+2y^2}{x+y}.\)
Chúc bạn học tốt!
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
