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Bài 1:
a) \(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot2\cdot1=25-8=17\)
Vì Δ>0 nên phương trình \(2x^2-5x+1=0\) có hai nghiệm là:
\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{5-\sqrt{17}}{2\cdot2}=\frac{5-\sqrt{17}}{4}\\x_2=\frac{5+\sqrt{17}}{2\cdot2}=\frac{5+\sqrt{17}}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5-\sqrt{17}}{4};\frac{5+\sqrt{17}}{4}\right\}\)
b) Ta có: \(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(S=\left\{\frac{-1}{2}\right\}\)
c) Ta có: \(-3x^2+2x+8=0\)
\(\Leftrightarrow-3x^2+6x-4x+8=0\)
\(\Leftrightarrow-3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\-3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{-4}{3}\right\}\)
d) Ta có: \(5x^2-6x-1=0\)
\(\Delta=b^2-4\cdot a\cdot c=\left(-6\right)^2-4\cdot5\cdot\left(-1\right)=56\)
Vì Δ>0 nên phương trình \(5x^2-6x-1=0\) có hai nghiệm là:
\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{6-\sqrt{56}}{2\cdot5}=\frac{3-\sqrt{14}}{5}\\x_2=\frac{6+\sqrt{56}}{2\cdot5}=\frac{3+\sqrt{14}}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3-\sqrt{14}}{5};\frac{3+\sqrt{14}}{5}\right\}\)
e) Ta có: \(-3x^2+14x-8=0\)
\(\Leftrightarrow-3x^2+12x+2x-8=0\)
\(\Leftrightarrow-3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(-3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\-3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\frac{2}{3}\right\}\)
g) Ta có: \(-7x^2+4x-3=0\)
\(\Delta=b^2-4ac=4^2-4\cdot\left(-7\right)\cdot\left(-3\right)=-68\)
Vì Δ<0 nên phương trình \(-7x^2+4x-3=0\) không có nghiệm
Vậy: S=∅
1)\(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow\hept{\begin{cases}7\ge0\\x-1=49\end{cases}\Leftrightarrow x=50}\)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
a) \(\sqrt{3}x-\sqrt{12}=0< =>\sqrt{3}x=\sqrt{12}=>x=2\)
Vay S = { 2 }
b) \(\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}< =>\sqrt{2}x=\sqrt{8}+\sqrt{18}-\sqrt{2}< =>\sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}\) <=> \(\sqrt{2}x=4\sqrt{2}=>x=4\)
Vay S = { 4 }
c) \(\sqrt{5}x^2-\sqrt{20}=0< =>\sqrt{5}x^2=\sqrt{20}< =>x^2=2=>x=\sqrt{2}\)
Vay S = {\(\sqrt{2}\) }
d) \(\sqrt{x^2+6x+9}=3x+6< =>\sqrt{\left(x+3\right)^2}=3x+6< =>x+3=3x+6< =>-2x=\) \(3=>x=-\dfrac{3}{2}\)
Vay S = { - 3/2 }
e) \(\sqrt{x^2-4x+4}-2x+5=0< =>\sqrt{\left(x-2\right)^2}-2x+5=0< =>x-2-2x+5=0\) <=> \(-x+3=0< =>-x=-3=>x=3\)
Vay S = { 3 }
F) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
<=> \(\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
Vay S = { 1/2 }
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả