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a)\(\sqrt{2x^2+x+6}+\sqrt{x^2+x+2}=x+\dfrac{4}{x}\)
\(pt\Leftrightarrow\sqrt{2x^2+x+6}-3+\sqrt{x^2+x+2}-2=x+\dfrac{4}{x}-5\)
Liên hợp quy đồng nốt
\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)
\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
-3x+3=\(2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)
Chắc tới đây bạn làm đc rồi nhỉ
\(4x^2-4-3x=\sqrt[3]{x^2\left(x^2-1\right)}\)
\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)
dat \(\left(x-1\right)\left(x+1\right)=y\)
\(4y-3x=\sqrt[3]{x^2y}\)
\(\Leftrightarrow\left(4y-3x\right)^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)
\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)
\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)
de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)
\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)
câu b tương tự nhé bạn
b. \(\sqrt{x-4}+\sqrt{x^2-3x+4}=x\)
(ĐKXĐ: \(x\ge4\))
\(\Leftrightarrow\sqrt{x^2-3x+4}=x-\sqrt{x-4}\)
\(\Leftrightarrow x^2-3x+4=x^2+x-4-2\sqrt{x\left(x-4\right)}\)
\(\Leftrightarrow x^2-3x+4-x^2-x+4+2\sqrt{x^2-4x}=0\Leftrightarrow-4x+8+2\sqrt{x^2-4x}=0\Leftrightarrow-2\left(2x-4-\sqrt{x^2-4x}\right)=0\Leftrightarrow2x-4-\sqrt{x^2-4x}=0\Leftrightarrow\sqrt{x^2-4x}=2x-4\Leftrightarrow x^2-4x=4x^2+16-16x\Leftrightarrow x^2-4x^2-4x+16x-16=0\Leftrightarrow-3x^2+12x-16=0\Leftrightarrow3x^2-12x+16=0\)
Ta có: \(\Delta=b^2-4ac=\left(-12\right)^2-4.3.16=-48< 0\)
=> pt vô nghiệm.
Vậy pt đã cho vô nghiệm.
ĐK : x > 3/2
Đặt \(\sqrt{3x-2}=a\left(a>0\right)\) . Khi đó pt thành :
\(1+\dfrac{x}{a}=\dfrac{1+a}{x}\Leftrightarrow\dfrac{a+x}{a}=\dfrac{a+1}{x}\Leftrightarrow a^2+a=ax+x^2\Leftrightarrow x^2+a\left(x-1\right)-a^2=0\)
hay \(\sqrt{3x-2}\left(x-1\right)+x^2-3x+2=0\Leftrightarrow\left(\sqrt{3x-2}-1\right)\left(x-1\right)+x^2-2x+1=0\Leftrightarrow\dfrac{3x-3}{\sqrt{3x-2}+1}\left(x-1\right)+\left(x-1\right)^2=0\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x-2}+1}+\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x-2}+1}+1\right)=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
Vì \(\dfrac{3}{\sqrt{3x-2}+1}+1>0\)
Vậy nghiệm của pt là x = 1