giải phương trình
a, \(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)
b, \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải bất phương trình :
a, \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}\dfrac{< }{ }5\sqrt{x+1}\)
b, \(2x\sqrt{x}+\dfrac{5-4x}{\sqrt{x}}\dfrac{>}{ }\sqrt{x+\dfrac{10}{x}-2}\)
c, \(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8< 0\)
Giải phương trình:
1, \(4\left(2x^2+1\right)+3\left(x^2-2x\right)\sqrt{2x-1}=2\left(x^3+5x\right)\)
2, \(\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}\)
3, \(\sqrt{5x^2-14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
Giải phương trình:
a) \(2\left(1-x\right)\sqrt{2x^2+2x-1}=x^2-x+1\)
b) \(\sqrt{x-1}+x-1=\sqrt{2\left(x-3\right)^2+2x-2}\)
c) \(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)
d) \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
m làm phần 10 chưa
T làm được phần c rồi
Trao đổi đê
Giải phương trình:
1, \(x^2\sqrt{x}+\left(x-5\right)^2\sqrt{5-x}=11\left(\sqrt{x}+\sqrt{5-x}\right)\)
2, \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+3}=0\)
3, \(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=2\sqrt{2}\)
4, \(\sqrt{x^2-\dfrac{1}{4x}}+\sqrt{x-\dfrac{1}{4x}}=x\)
5, \(\sqrt{5x^2+14x+9}-\sqrt{x^2-1-20}=5\sqrt{x+1}\)
giải phương trình
\(x^2+3\sqrt{x^2-1}=\sqrt{x^2-x+1}\)
\(\sqrt{x^2+2x}+\sqrt{2x+1}=\sqrt{3x^2+4x+1}\)
\(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
\(\sqrt{5x^2-14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
Giải phương trình:
a) \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
b)\(x+\sqrt{x+4}=\sqrt{2x^2-10x+17}+3\)
b, ĐK \(x\ge-4\)
PT
<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)
<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)
Giải (2)
=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)
<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)
<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)
<=> \(x^2-7x-4=6\sqrt{x+4}\)
<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)
Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)
=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)
=> \(a^2-b^2+6\left(a-b\right)=0\)
<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)
+ a=b
=> \(x-6=\sqrt{x+4}\)
=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)
+ a+b+6=0
=> \(x+\sqrt{x+4}=0\)(loại)
Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)
Giải các pt
A.\(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)
B.\(\sqrt{5x^2-14x+9}-\sqrt{x^2-x-20}=5\sqrt{x-1}\)
C.\(\sqrt{x^2-3x-2}+\sqrt{x+3}=\sqrt{x+2}+\sqrt{x^2+2x-3}\)
Mọi người giúp mình với <3<3<3<3<3thank you!!!
giải pt
a) \(\sqrt[3]{x+6}+\sqrt{x-1}=x^2-1\)
b) \(\sqrt[3]{x-9}+2x^2+3x=\sqrt{5x-1}+1\)
c) \(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
d) \(\sqrt{x+1}-2\sqrt{4-x}=\frac{5\left(x-3\right)}{\sqrt{2x^2+18}}\)
e) \(x^3+5x^2+6x=\left(x+2\right)\left(\sqrt{2x+2}+\sqrt{5-x}\right)\)