If \(\dfrac{x^2-4}{x+2}=\dfrac{3x}{2}\) then x = \(\left\{.....................\right\}\)
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Y
26 tháng 11 2023
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
22 tháng 12 2020
a) Ta có: \(\dfrac{9-3x}{x^2+3x+4}-\dfrac{3x-23}{\left(1-x\right)\left(x+4\right)}\)
\(=\dfrac{9-3x}{x^2+3x+4}+\dfrac{3x-23}{x^2+3x-4}\)
\(=\dfrac{\left(9-3x\right)\left(x^2+3x-4\right)}{\left(x^2+3x+4\right)\left(x^2+3x-4\right)}+\dfrac{\left(3x-23\right)\left(x^2+3x+4\right)}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
\(=\dfrac{9x^2+27x-36-3x^3-9x^2+12x+3x^3+9x^2+12x-23x^2-69x-92}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
\(=\dfrac{-14x^2-18x-128}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)
b) Ta có: \(\dfrac{4-x}{x^3+2x}-\dfrac{x+5}{x^3-x^2+2x-2}\)
\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\dfrac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\dfrac{-2}{x\left(x-1\right)}\)
25 tháng 2 2023
a: =>x^2+4x-4x+1=0
=>x^2+1=0
=>Loại
b: =>2x-6+4=2x+2
=>-2=2(loại)
c: =>2(x+3)-2x-1=1
=>6-1=1
=>5=1(loại)
d =>x+3=0
=>x=-3(loại)
e: =>x^2-3x^2+3x-3x-2=0
=>-2x^2-2=0
=>x^2+1=0
=>Loại
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
31 tháng 1 2021
a) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+5x=30-144\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: S={6}
b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)
\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)
\(\Leftrightarrow2-10x=-12x+12\)
\(\Leftrightarrow2-10x+12x-12=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
hay x=5
Vậy: S={5}
c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow6-2x-8=5x+10\)
\(\Leftrightarrow-2x+2-5x-10=0\)
\(\Leftrightarrow-7x-8=0\)
\(\Leftrightarrow-7x=8\)
hay \(x=-\dfrac{8}{7}\)
Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)
d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)
\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)
\(\Leftrightarrow35-15x-2x-10-10=0\)
\(\Leftrightarrow-17x+15=0\)
\(\Leftrightarrow-17x=-15\)
hay \(x=\dfrac{15}{17}\)
Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)
ND
1 tháng 2 2021
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2019
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2019
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
Condition \(x\ne-2\)
We have :
\(\dfrac{x^2-4}{x+2}=\dfrac{3x}{2}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+2\right)}{x+2}=\dfrac{3x}{2}\)
\(\Leftrightarrow x-2=\dfrac{3x}{2}\)
\(\Leftrightarrow2\left(x-2\right)=3x\)
\(\Leftrightarrow2x-4=3x\)
\(\Leftrightarrow x=-4\)
So the value of x is : \(-4\)
Thanks!