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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)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
giải pt
a)\(\dfrac{1}{x+1}+\dfrac{3}{2x+1}=\dfrac{8}{x-2}\)
b)\(\sqrt{2x+1}+\sqrt{3-x}=\sqrt{3x+5}\)
a:
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
=>|x-3|=3
=>x-3=3 hoặc x-3=-3
=>x=0 hoặc x=6
b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
=>\(\left|\sqrt{x-1}+1\right|=2\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)
=>x-1=1
=>x=2
c:
ĐKXĐ: x>4/5
PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)
=>\(\dfrac{5x-4}{x+2}=4\)
=>5x-4=4x+8
=>x=12(nhận)
d: ĐKXĐ: x-4>=0 và x+1>=0
=>x>=4
PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)
=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)
=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)
=>\(\sqrt{x^2-3x-4}=14-x\)
=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196
=>x<=14 và -3x-4=-28x+196
=>x<=14 và 25x=200
=>x=8(nhận)
a) \(\sqrt{x^2-6x+9}=3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
\(\Leftrightarrow\left|x-3\right|=3 \)
TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)
Pt trở thành:
\(x-3=3\) (ĐK: \(x\ge3\))
\(\Leftrightarrow x=3+3\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)
Pt trở thành:
\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=-3+3\)
\(\Leftrightarrow x=0\left(tm\right)\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4-x\)
\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)
\(\Leftrightarrow4x-4=16-8x+x^2\)
\(\Leftrightarrow x^2-12x+20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))
\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)
\(\Leftrightarrow5x-4=4x+8\)
\(\Leftrightarrow x=12\left(tm\right)\)
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)
\(\Rightarrow\) vô nghiệm
b) ĐK: \(x\ne1;-1\)
\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)
\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)
a) Ta có: \(\Delta'=100-104=-4< 0\)
Vậy phương trình vô nghiệm.
b) ĐKXĐ: \(x\ne1;x\ne-1\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x-19=45x+45\)
\(\Leftrightarrow43x=-64\)
\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)
Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)