\(x^2-5x-2\sqrt{3x}+12=0\) giai pt
Giai pt \(x^2-5x-2\sqrt{3x}+12=0\)
ĐKXĐ \(x\ge0\)
\(x^2-5x-2\sqrt{3x}+12=0\)
\(\Rightarrow x^2-6x+9+x-2\sqrt{3x}+3=0\)
\(\Rightarrow\left(x-3\right)^2+\left(\sqrt{x}-\sqrt{3}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^2=0\\\left(\sqrt{x}-\sqrt{3}\right)^2=0\end{cases}}\Leftrightarrow x=3\)
Vậy...
giai pt \(\sqrt{2x^2^{ }+5x+12}+\sqrt{2x^2+3x+2}=x+5\)
\(y=\frac{1}{x^2+\sqrt{x}}\left(nvb\int^{42}_{3^{2_{1\frac{12}{23}}}}\right)\)
GIẢI CÁC PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
\(\sqrt{x^2-2x+6}=2x-3\)
\(\sqrt{3x^2-2x+6}+3-2x=0\)
Giai pt :
\(7\sqrt{4x^2+5x-1}-14\sqrt{x^2-3x+3}=17x-13\)
Giai pt sau:
1/x^2-3x+2 +1/x^2-5x+6 +1/x^2-7x+12 =2(Tất cả =2 nhé!)
=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)
=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)
=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)
=>2x^2-10x+8=3
=>2x^2-10x+5=0
=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)
giai pt sau
\(\sqrt{3x-1}-\sqrt{x+2}.\sqrt{3x^2+7x+2}+4=4x-2\)
\(x^2-5x+3.\sqrt{2x-1}=2.\sqrt{14-2x}+5\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
nhiều thế giải ko đổi đâu bạn
đkxđ : \(\frac{1}{2}\le x\le7\)
\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)
\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)
\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)
\(\Leftrightarrow x=5\)
còn bài a,c lười đánh lắm
Giai pt:
a. \(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
b. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
a/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)
\(\Leftrightarrow x-1=8x-3+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
Do \(x\ge1\Rightarrow2-7x< 0\Rightarrow\left\{{}\begin{matrix}VP\ge0\\VT< 0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|=2\)
Mà \(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)
Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{x-1}\ge0\Rightarrow x\le2\Rightarrow1\le x\le2\)
Vậy nghiệm của pt là \(1\le x\le2\)
GIẢI PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
a, ĐKXĐ:...
\(\sqrt{5x+10}=8-x\\ \Leftrightarrow5x+10=64-16x+x^2\\ \Leftrightarrow x^2-21x+54=0\)
.....
b, ĐKXĐ:...
\(\sqrt{4x^2+x-12}=3x-5\\ \Leftrightarrow4x^2+x-12=9x^2-30x+25\\ \Leftrightarrow5x^2-31x+37=0\)
.....
Giai pt:
a. \(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
b. \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)