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15 tháng 1 2019
\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{2x^2+4}{x^2-4}=\frac{2x^2+4}{x^2-4}\)
Vậy phương trình này có vô số nghiệm x thỏa mãn trừ x khác 2 và -2
20 tháng 8 2021
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
PN
1 tháng 8 2020
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
PN
1 tháng 8 2020
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
30 tháng 9 2021
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
NH
21 tháng 1 2018
2. \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow\)\(x^2+9x+x+9=x^2+5x+3x+15\)
\(\Leftrightarrow x^2+9x+x-x^2-5x-3x=15-9\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\Rightarrow x=3\)
\(S=\left\{3\right\}\)
21 tháng 1 2018
\(1,5-\left(6-x\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow x+8x=12-5+6\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\dfrac{13}{9}\)
Vậy tập nghiệm của pt là \(S=\left\{\dfrac{13}{9}\right\}\)
\(2,\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)
\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của pt là S = { 3 }
\(3,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\dfrac{9\left(5x-2\right)-24}{12}=\dfrac{28x-60\left(x-7\right)}{12}\)
\(\Rightarrow45x-18-24=28x-60x+420\)
\(\Leftrightarrow45x-28x+60x=420+18+24\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm của pt là S = { 6 }
\(4,3\left(x+1\right)\left(2x+5\right)=3\left(x+1\right)\left(7x-4\right)\)
\(\Leftrightarrow3\left(x+1\right)\left(2x+5\right)-3\left(x+1\right)\left(7x-4\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(2x+5-7x+4\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(-5x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\-5x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{9}{5}\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{-1;\dfrac{9}{5}\right\}\)
\(5,\left(x-2\right)^2-\left(3x+1\right)^2+x\left(4x-1\right)=0\)
\(\Leftrightarrow\left(x-2-3x-1\right)\left(x-2+3x+1\right)+x\left(4x-1\right)=0\)
\(\Leftrightarrow\left(-2x-3\right)\left(4x-1\right)+x\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-2x-3+x\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(-x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{\dfrac{1}{4};-3\right\}\)
9 tháng 2 2021
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
11 tháng 3 2023
`a,4x-10=0 `
`<=> 4x=10`
`<=>x=10/4`
`<=>x=5/2`
`b, 7-3x=9-x `
`<=>-3x+x=9-7`
`<=>-2x=2`
`<=>x=-1`
`c, 2x-(3-5x) = 4(x+3)`
`<=>2x-3+5x=4x+12`
`<=>2x+5x-4x=12+3`
`<=>3x=15`
`<=>x=5`
`d, 5-(6-x)=4(3-2x) `
`<=>5-6+x=12-8x`
`<=>x+8x=12-5+6`
`<=>9x=13`
`<=>x=13/9`
`e, 4(x+3)=-7x+17 `
`<=>4x+12=-7x+17`
`<=>4x+7x=17-12`
`<=>11x=5`
`<=>x=5/11`
`f, 5(x-3) - 4=2(x-1)+7`
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`g, 5(x-3)-4=2(x-1)+7 `
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`h,4(3x-2)-3(x-4)=7x+20`
`<=>12x-8-3x+12=7x+20`
`<=>12x-3x-7x=20+8+12`
`<=>2x=40`
`<=>x=20`
\(\frac{-7x^2+4}{x^3+1}=\frac{5}{x^2-x+1}-\frac{1}{1+x}\)
ĐKXĐ : x ≠ -1
pt <=> \(\frac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=0\)
<=> \(\frac{-7x^2+4-5x-5+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=0\)
<=> \(\frac{-6x^2-6x}{\left(x+1\right)\left(x^2-x+1\right)}=0\)
<=> \(\frac{-6x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\)
<=> \(\frac{-6x}{x^2-x+1}=0\)
=> -6x = 0 <=> x = 0 ( tm )
Vậy phương trình có nghiệm x = 0