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a) 5.(x^2-3x+1)+x.(1-5x)=x-2
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-14x-x=-2-5\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=\frac{7}{15}\)
b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)
\(\Leftrightarrow-4x^2+15x-10=0\)
Đề sai???
\(c,12x^2-4x\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)
\(\Leftrightarrow10x=-17\)
\(\Leftrightarrow x=-\frac{17}{10}\)
\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{3}{2}\)
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)
=>-8x^2+11x-10=0
=>\(x\in\varnothing\)
2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
=>-14x+5=x-2
=>-15x=-7
=>x=7/15
3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)
=>10x=-17
=>x=-17/10
4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)
=>18x+3=7x-3
=>11x=-6
=>x=-6/11
5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)
\(\Leftrightarrow3x^2+2x+10-4+x=0\)
=>3x^2+3x+6=0
hay \(x\in\varnothing\)
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
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a) ĐKXĐ: \(x\ne1\)
Ta có: \(\frac{7x-3}{x-1}=\frac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
hay \(x=\frac{7}{19}\)
Vậy: \(x=\frac{7}{19}\)
b) ĐKXĐ: \(x\ne-1\)
Ta có: \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x-1-x=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow29x=11\)
hay \(x=\frac{11}{29}\)
Vậy: \(x=\frac{11}{29}\)
c) ĐKXĐ: \(x\notin\left\{\frac{-2}{3};\frac{1}{3}\right\}\)
Ta có: \(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow15x^2-8x+1-15x^2+11x+14=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
d) ĐKXĐ: \(x\notin\left\{1;\frac{-4}{3}\right\}\)
Ta có: \(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow12x^2+37x+28-12x^2+7x+5=0\)
\(\Leftrightarrow44x+33=0\)
\(\Leftrightarrow44x=-33\)
hay \(x=\frac{-3}{4}\)
Vậy: \(x=\frac{-3}{4}\)
a)
\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{21x-9-2x+2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow\frac{19x-7}{3\cdot\left(x-1\right)}=0\\ \Rightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)
b)
\(\frac{2\cdot\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{12-28x}{2\cdot\left(1+x\right)}-\frac{1+x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{12-28x-1-x}{2\cdot\left(1+x\right)}=0\\ \Leftrightarrow\frac{11-29x}{2\cdot\left(1+x\right)}=0\\\Rightarrow11-29x=0\\ \Rightarrow x=\frac{11}{29}\)
c)
\(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{15x^2-8x+1}{\left(3x+2\right)\cdot\left(3x-1\right)}-\frac{15x^2-11x-14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{15x^2-8x+1-15x^2+11x+14}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Leftrightarrow\frac{3x+15}{\left(3x+2\right)\cdot\left(3x-1\right)}=0\\ \Rightarrow3x+15=0\\ \Rightarrow x=-5\)
d)
\(\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\ \Leftrightarrow\frac{12x^2+37x+28}{\left(x-1\right)\cdot\left(3x+4\right)}-\frac{12x^2-7x-5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{12x^2+37x+28-12x^2+7x+5}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow\frac{44x+33}{\left(x-1\right)\cdot\left(3x+4\right)}=0\\ \Leftrightarrow44x+33=0\\ \Rightarrow x=-\frac{3}{4}\)
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
\(a,5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2-x+2=0\)
\(\Leftrightarrow-15x+7=0\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=-\frac{7}{-15}\)
\(\Leftrightarrow x=\frac{7}{15}\)