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21 tháng 2 2023
a: \(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)
=>-2x^2+x+1-2x^2+2x=0
=>-4x^2+3x+1=0
=>4x^2-3x-1=0
=>4x^2-4x+x-1=0
=>(x-1)(4x+1)=0
=>x=1(loại) hoặc x=-1/4(nhận)
b: \(\Leftrightarrow\dfrac{440}{x-2}-\dfrac{440}{x}=1\)
=>x(x-2)=440x-440x+880
=>x^2-2x-880=0
=>\(x=1\pm\sqrt{881}\)
c: \(\Leftrightarrow\dfrac{x+5+x}{x\left(x+5\right)}=\dfrac{1}{6}\)
=>x^2+5x=6(2x+5)
=>x^2+5x-12x-30=0
=>x^2-7x-30=0
=>(x-10)(x+3)=0
=>x=10 hoặc x=-3
d: =>(x-1)(x+1)-x=2x-1
=>x^2-1-x=2x-1
=>x^2-x-2x=0
=>x(x-3)=0
=>x=0(loại) hoặc x=3(nhận)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
26 tháng 11 2023
a: ĐKXĐ: x<>-1 và y<>-1
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+2-2}{x+1}+\dfrac{y+1-1}{y+1}=2\\\dfrac{x+1-1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{2}{x+1}+1-\dfrac{1}{y+1}=2\\1-\dfrac{1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=2-3=-1\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=1+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=-1\\\dfrac{2}{x+1}-\dfrac{6}{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y-1}=3\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=-\dfrac{7}{3}\\\dfrac{1}{x+1}-3:\dfrac{-7}{3}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}+3\cdot\dfrac{3}{7}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}=2-\dfrac{9}{7}=\dfrac{5}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x+1=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: y<>0 và y<>-12
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y}-\dfrac{x}{y+12}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\cdot\dfrac{x}{y+12}=3\left(vôlý\right)\\\dfrac{x}{y}-\dfrac{x}{y+12}=1\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\varnothing\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)\)
13 tháng 4 2023
Δ=(2m-2)^2-4(m+1)
=4m^2-8m+4-4m-4
=4m^2-12m
Để phương trình co hai nghiệm thì 4m^2-12m>0
=>m>3 hoặc m<0
x1/x2+x2/x1=4
=>x1^2+x2^2=4x1x2
=>(x1+x2)^2-2x1x2=4x1x2
=>(2m-2)^2-6(m+1)=0
=>4m^2-8m+4-6m-6=0
=>4m^2-14m-2=0
=>\(m=\dfrac{7\pm\sqrt{57}}{2}\)
14 tháng 11 2018
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}-\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
20 tháng 11 2022
=>12/(x+y-1)-15/(2x-y+3)=15/2 và 12/(x+y-1)-4/(2x-y+3)=28/5
=>x+y-1=22/9; 2x-y+3=-110/19
=>x+y=31/9; 2x-y=-167/19
=>x=-914/513; y=2681/513
31 tháng 1 2023
Đặt : \(a=1-\dfrac{2x-1}{x+1},ta\) có :
\(\dfrac{12\left(2x-1\right)}{x+1}-20=-12\left(1-\dfrac{2x-1}{x+1}\right)-8=-12a-8\)
Do đó pt có dạng : \(a^3+6a^2=-12a-8\)
\(\Leftrightarrow\left(a-2\right)^3=0\)
\(\Leftrightarrow a=-2\)
vậy pt đã cho tương đương với :
\(1-\dfrac{2x-1}{x+1}=2\) hay \(\dfrac{2x-1}{x+1}=3\)
\(\Leftrightarrow2x-1=3\left(x+1\right)\)
\(\Leftrightarrow x=-4\)
Vậy ...
31 tháng 1 2023
Đặt (2x-1)/(x+1) = a, để rút gọn vế giải cho dễ
(1-a)^3 + 6a = 12a - 20
Tìm a, từ đó giải ra x
ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2\left(25-x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne25\end{matrix}\right.\)
\(\dfrac{1}{2x}+\dfrac{1}{2\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25-x+x}{2x\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25}{-2x^2+50x}=\dfrac{1}{12}\\ \Leftrightarrow-2x^2+50x=300\\ \Leftrightarrow-2x^2+50x-300=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\left(tm\right)\\x=10\left(tm\right)\end{matrix}\right.\)
Vậy...