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NT
1
D
6 tháng 8 2019
\(ĐK:x\ne0;x\ne-6\)
⇔ \(\frac{720\left(x+6\right)}{6x\left(x+6\right)}=\frac{6x\left(x+6\right)}{6x\left(x+6\right)}+\frac{x\left(x+6\right)}{6x\left(x+6\right)}+\frac{6x\left(120-x\right)}{6x\left(x+6\right)}\)
\(\Rightarrow720x+4320=6x^2+36x+x^2+6x+720x-6x^2\)
\(\Leftrightarrow6x^2+36x+x^2+6x+720x-6x^2-720x-4320=0\)
\(\Leftrightarrow x^2+42x-4320=0\)
\(\Leftrightarrow x^2+90x-48x-4320=0\)
\(\Leftrightarrow\left(x+90\right)\left(x-48\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+90=0\\x-48=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-90\\x=48\end{matrix}\right.\) ( tm )
NT
1
DA
27 tháng 2 2015
câu a
x/3 +20 =x/2
x/2 - x/3 = 20
(3x-2x)/6 = 20
x/6 = 20
x = 20*6
x=120
câu b
x/(x-1) + 2x/x*x = 0 (x khác 0 ,1)
(x*x*x + 2x *(x-1)) / (x-1) * x*x = 0
x*x*x + 2*x*x - 2*x = 0
x*(x*x + 2*x -2 ) =0
x=0 hoặc x*x+2*x-2=0
x=0 hoặc (x*x + 2x + 1)-3 =0
x=0 hoặc (x + 1)*(x+1)=3
x=0 hoặc x+1 = căn 3 hoặc x=âm căn3
x=0 hoặc x =căn 3 trừ 1 hoặc x = âm căn 3 trừ một
2 tháng 2 2019
a,\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x+1}.\frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)\(\Leftrightarrow\left(\frac{x^2-x+x^2+x}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0\)\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}-10\right)\left(\frac{2x^2}{x^2-1}+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2}{x^2-1}=10\\\frac{2x^2}{x^2-1}=-9\end{cases}\Leftrightarrow......}\)
b,Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\Rightarrow ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\)
Từ đó ta có phương trình:\(20a^2-5b^2+48ab=0\Leftrightarrow20a^2-2ab-5b^2+50ab=0\)
\(\Leftrightarrow2a\left(10a-b\right)+5b\left(10a-b\right)=0\Leftrightarrow\left(2a+5b\right)\left(10a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=-5b\\10a=b\end{cases}}\)
TH1:\(2a=-5b\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\)\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\)\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\)\(\Leftrightarrow7x^2+9x+14=0\)
\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\Leftrightarrow7\left(x^2+2.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{28}=0\)
\(\Leftrightarrow7\left(x+\frac{9}{14}\right)^2+\frac{311}{28}=0\),vô lí
TH2:Tự làm nhé ,tương tự
BL
27 tháng 2 2020
\(a.\frac{7x-3}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{7x-3}{x-1}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{2\left(7x-3\right)}{2.\left(x-1\right)}-\frac{3\left(x-1\right)}{2\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{14x-6-3x+3}{2\left(x-1\right)}=0\)
\(\Leftrightarrow11x-3=0\)
\(\Leftrightarrow x=\frac{3}{11}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6-14x}{1+x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{2\left(6-14x\right)}{2\left(1+x\right)}-\frac{1+x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow\frac{12-28x-1-x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow x=\frac{11}{29}\)
\(c.\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}-\frac{3-x}{x-2}=0\)
\(\Leftrightarrow\frac{1+3x-6-3+x}{x-2}=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
\(d.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\)
\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25-20}{x^2-25}=0\)
\(\Leftrightarrow20x-20=0\)
\(\Leftrightarrow x=10\)
21 tháng 4 2019
\(c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.\)
\(\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0\)
\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
=>\(\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}\)
Vậy.......
6 tháng 2 2017
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
\(\Leftrightarrow\frac{200\left(x+20\right)}{2x\left(x+20\right)}-\frac{240x}{2x\left(x+20\right)}=\frac{x\left(x+20\right)}{2x\left(x+20\right)}\) đk: x\(\ne0\) , x \(\ne-20\)
\(\Rightarrow200x+4000-240x=x^2+20x\)
\(\Leftrightarrow-x^2-60x+4000=0\)
\(\Leftrightarrow x^2+60x-4000=0\)
\(\Leftrightarrow x^2+100x-40x-4000=0\)
\(\Leftrightarrow\left(x+100\right)\left(x-40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+100=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-100\left(tmđk\right)\\x=40\left(tmđk\right)\end{matrix}\right.\)
Vậy S\(=\left\{-100;40\right\}\)
\(\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2}\)
\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2},x\ne0,x\ne-20\)
\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{200\left(x+20\right)-240x-x\left(x+20\right)}{2x\left(x+20\right)}=0\)
\(\Leftrightarrow\frac{200x+4000-240x-x^2-20x}{2x\left(x+20\right)}=0\)
\(\Leftrightarrow-60x+4000-x^2=0\)
\(\Leftrightarrow-x^2-60x+4000=0\)
\(\Leftrightarrow x^2+60x-4000=0\)
\(\Leftrightarrow\frac{-60\pm\sqrt{60^2}-4.1\left(-4000\right)}{2}\)
\(\Leftrightarrow\frac{-60\pm\sqrt{3600+16000}}{2}\)
\(\Leftrightarrow\frac{-60\pm\sqrt{19600}}{2}\)
\(\Leftrightarrow\frac{-60\pm140}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-60+140}{2}\\\frac{-60-140}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-100\end{matrix}\right.,x\ne0,x\ne-20\)