giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải các PT sau :
a) \(\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4\)
b) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
d) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)
e) \(\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16\)
f)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
giải pt :
a, \(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
b, \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^2+2\)
c, \(\sqrt{4x^2+x+6}=4x-2+7\sqrt{x+1}\)
d, \(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)
giải pt :
a, \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^2+2\)
b, \(\sqrt{4x^2+x+6}=4x-2+7\sqrt{x+1}\)
c, \(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)
giải pt :
a, \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
b, \(2\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0\)
c, \(\left(x+3\right)\sqrt{-x^2-8x+48}=x-24\)
d, \(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)\left(2-x\right)}=3\)
e, \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
giải pt
a) \(x+\sqrt{4-x^2}-3x\sqrt{4-x^2}=2\)
b) \(2\left(\sqrt{4-x^2}+4\right)\left(\sqrt{x+2}+\sqrt{2-x}\right)-5=0\)
c) \(\left(\sqrt{x^2-4}-x+1\right)\left(\sqrt{x-2}+\sqrt{x+2}\right)+2=0\)
d) \(\sqrt{x+2}-\sqrt{x-1}=\frac{6}{\sqrt{x^2+x-2}-x}\)
e) \(\frac{2}{\sqrt{x-1}+\sqrt{3-x}}=1+\sqrt{3+2x-x^2}\)
a/ ĐKXĐ: \(-2\le x\le2\)
Đặt \(x+\sqrt{4-x^2}=a\Rightarrow a^2=4+2x\sqrt{4-x^2}\Rightarrow x\sqrt{4-x^2}=\frac{a^2-4}{2}\)
\(\Rightarrow a-\frac{3\left(a^2-4\right)}{2}=2\)
\(\Leftrightarrow-3a^2+2a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{4-x^2}=2\\x+\sqrt{4-x^2}=-\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{4-x^2}=2-x\\3\sqrt{4-x^2}=-4-3x\left(x\le-\frac{4}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x^2=x^2-4x+4\\12\left(4-x^2\right)=9x^2+24x+16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-4x=0\\21x^2+24x-32=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\frac{-12+4\sqrt{51}}{2}\left(l\right)\\x=\frac{-12-4\sqrt{51}}{2}\end{matrix}\right.\)
Mấy câu còn lại và bài kia tầm 30ph nữa sẽ làm, bận chút xíu việc
b/ ĐKXĐ: \(-2\le x\le2\)
\(\Leftrightarrow\left(2\sqrt{4-x^2}+4+4\right)\left(\sqrt{x+2}+\sqrt{2-x}\right)-5=0\)
Đặt \(\sqrt{x+2}+\sqrt{2-x}=a>0\Rightarrow a^2=4+2\sqrt{4-x^2}\)
\(\Rightarrow\left(a^2+4\right)a-5=0\)
\(\Leftrightarrow a^3+4a-5=0\Leftrightarrow\left(a-1\right)\left(a^2+a+5\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{x+2}+\sqrt{2-x}=1\)
\(\Leftrightarrow4+2\sqrt{4-x^2}=1\Rightarrow2\sqrt{4-x^2}=-3\)
Vậy pt vô nghiệm
Thật ra bài này có thể biện luận vô nghiệm ngay từ đầu:
\(\sqrt{x+2}+\sqrt{2-x}\ge\sqrt{x+2+2-x}=2\)
\(2\left(\sqrt{4-x^2}+4\right)\ge2.4=8\)
\(\Rightarrow VT>8.2-5=11>0\) nên pt vô nghiệm
c/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left(\sqrt{x^2-4}-x+1\right)\left(\sqrt{x-2}+\sqrt{x+2}\right)=-2\)
Do \(\sqrt{x+2}>\sqrt{x-2}\Rightarrow\sqrt{x+2}-\sqrt{x-2}\ne0\)
Nhân cả 2 vế của pt với \(\sqrt{x+2}-\sqrt{x-2}\) và rút gọn ta được:
\(4\left(\sqrt{x^2-4}-x+1\right)=-2\left(\sqrt{x+2}-\sqrt{x-2}\right)\)
Đặt \(\sqrt{x+2}-\sqrt{x-2}=a>0\)
\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\Rightarrow\sqrt{x^2-4}-x=-\frac{a^2}{2}\)
Phương trình trở thành:
\(4\left(-\frac{a^2}{2}+1\right)=-2a\)
\(\Leftrightarrow-a^2+2=-a\Leftrightarrow a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+2}-\sqrt{x-2}=2\Leftrightarrow\sqrt{x+2}=2+\sqrt{x-2}\)
\(\Leftrightarrow x+2=x+2+4\sqrt{x-2}\)
\(\Rightarrow4\sqrt{x-2}=0\Rightarrow x=2\)
giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
