Giải :
a) \(x^2-\sqrt{2}x+\sqrt{5}x-\sqrt{10}=0\)
b) \(\left(x^2-x\right)^2-5x^2+5x+4=0\)
c) \(5x+\sqrt{5x-x^2}=x^2+6\)
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
Bài 2 : Giải các phương trình sau
1 , \(x\left(x+5\right)=2\sqrt[3]{x^2+5x-2}-2\)
2 , \(\sqrt[3]{x+5}+\sqrt[3]{x+6}=\sqrt[3]{2x+11}\)
3 , \(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
4 , \(x^2-2x-8=4\sqrt{\left(4-x\right)\left(x+2\right)}\)
5 , \(x^2+5x+2+2\sqrt{x^2+5x+10}=0\)
6 , \(\sqrt{2x^2+3x-5}=x+1\)
7 , \(\left(x-1\right)\left(x-3\right)+3\sqrt{x^2-4x+5}-2=0\)
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
3/ ĐKXĐ: \(\left|x\right|\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{x-\sqrt{x^2-1}}=a>0\\\sqrt[4]{x+\sqrt{x^2-1}}=b>0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}ab=1\\a+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}ab=1\\a=2-b^2\end{matrix}\right.\)
\(\Rightarrow b\left(2-b^2\right)=1\Leftrightarrow b^3-2b+1=0\)
\(\Leftrightarrow\left(b-1\right)\left(b^2+b-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=1\\b^2+b-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}b=1\\b=\frac{-1+\sqrt{5}}{2}\\b=\frac{-1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[4]{x+\sqrt{x^2-1}}=1\\\sqrt[4]{x+\sqrt{x^2-1}}=\frac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x^2-1}=1\\x+\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}-x\left(vn\right)\end{matrix}\right.\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Giải phương trình :
a.\(x^2+5x^2-3=0\)
b.\(x^2-\left(2\sqrt{3}-1\right)x+4\sqrt{3}-6=0\)
c.\(x^2-6x+9=0\)
d.\(x^2-4\sqrt{3}x-4=0\)
c: \(\Leftrightarrow x-3=0\)
hay x=3
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Giải các phương trình:
a) \(\left(3x-1\right)\left(3x+1\right)=x\left(1+8\sqrt{x+1}\right)\)
b) \(5x^2-5x\sqrt{x^2+x+4}+2x+5=0\)
c) \(9x^2+8x+9=9\left(x+1\right)\sqrt{2x^2+1}\)
d) \(5x^2+2x+2=5x\sqrt{x^2+x+1}\)
e) \(5x^2+20x-12=5\left(x-2\right)\sqrt{3x^2+x}\)
a/ ĐXĐK: ...
\(\Leftrightarrow9x^2-1-x-8x\sqrt{x+1}=0\)
\(\Leftrightarrow x^2-x-1+8x\left(x-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x^2-x-1+\frac{8x\left(x^2-x-1\right)}{x+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\Rightarrow x=...\\\frac{-8x}{x+\sqrt{x+1}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-8x=x+\sqrt{x+1}\)
\(\Leftrightarrow-9x=\sqrt{x+1}\) (\(x\le0\))
\(\Leftrightarrow81x^2-x-1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1-5\sqrt{13}}{162}\\x=\frac{1+5\sqrt{13}}{162}>0\left(l\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow3x^2+2\left(x^2+x+1\right)-5x\sqrt{x^2+x+1}=0\)
Đặt \(\sqrt{x^2+x+1}=a\)
\(\Leftrightarrow3x^2-5ax+2a^2=0\)
\(\Leftrightarrow\left(x-a\right)\left(3x-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a\\3x=2a\end{matrix}\right.\) (\(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=x\\2\sqrt{x^2+x+1}=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=x^2\\2\left(x^2+x+1\right)=9x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\7x^2-2x-2=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1+\sqrt{15}}{7}\)
Bài 1: Giải phương trình
1) \(\sqrt{4x^2+12x+9}=2-x\left(vớix\le0\right)\)
2) \(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ( với \(x^2+5x+4>0\))
3) \(\sqrt{5x+1}=4\)
4) \(\sqrt{3-x}=7\)
Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ĐK:\(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)
\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)
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) \(x^3+5x^2+6x=\left(x+2\right)\left(\sqrt{2x+2}+\sqrt{5-x}\right)\)