x=\(\dfrac{7}{\sqrt[3]{26+15\sqrt{3}}}-\sqrt{26+15\sqrt{3}}\)
Tính Q=\(\dfrac{x^6+x^4+4x^2}{40\left(x^4+4x^2-144\right)}\)
cho x =\(\dfrac{7-4\sqrt{3}}{\sqrt[3]{26-15\sqrt{3}}}-\sqrt[3]{26+15\sqrt{3}}\)
tính A= \(\dfrac{x^6+x^4+4x^2}{40\left(x^4+4x^2-144\right)}\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
Giải phương trình:
1. \(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
2. \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
3. \(\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}{3}\)
4. \(x^2+1-\left(x+1\right)\sqrt{x^2-2x+3}=0\)
5. \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
6. \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt[]{x}}{x+\sqrt{x}}\right)\)
\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\)
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
RÚT GON
\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))
\(=\left(\dfrac{1}{\sqrt{x}}-\dfrac{x}{\sqrt{x}}\right)\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{-\sqrt{x}}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{-\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\left(\sqrt{x}+1\right)^2\)
c:
b;
Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
bài 1 : rút gọn các biểu thức sau .
a, \(\sqrt{4\left(a-3\right)^2}+2\sqrt{a^2+4a+4}\left(a< -2\right)\)
b, \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-2\right)^2}}+\dfrac{x^2-1}{x-3}\left(x< 3\right)\)
c, \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
bài 2 thực hiện phép tính :\
a, \(\sqrt{8-\sqrt[2]{7}}\times\sqrt{8+\sqrt[2]{7}}\)
b, \(\sqrt{4+\sqrt{8}+}+\sqrt{2}+\sqrt{2+\sqrt{2}}\times\sqrt{2-\sqrt{2+2}}\)
c, \(\left(4+\sqrt{15}\right)\times\sqrt{10}-\sqrt{6}\times\sqrt{4-\sqrt{15}}\)
d, \(\left(2+\sqrt{3}\right)^2-\left(2-\sqrt{3}\right)\times\left(2+\sqrt{3}\right)\)
Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
Tính GTBT
a,M=\(\left(3x^3-x^2-1\right)^{2018}\) biết x = \(\dfrac{\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt[]{3}\right)}{\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}}\)
b,\(x^3+ax+b\) biết x=\(\sqrt[3]{\dfrac{-b}{2}+\sqrt{\dfrac{b^2}{4}+\dfrac{a^3}{27}}}+\sqrt[3]{\dfrac{-b}{2}-\sqrt{\dfrac{b^2}{4}+\dfrac{a^3}{27}}}\)
a)\(\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\left(9x+5\right)\)
b)\(\sqrt{9-4\sqrt{5}}\)
c)\(\sqrt{9-6x+4x^2}\)=4
d)\(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\)
b)\(\sqrt{9-4\sqrt{5}}\)=\(\sqrt{9-\sqrt{80}}\)=\(\sqrt{\dfrac{9+\sqrt{9^2-80}}{2}}-\sqrt{\dfrac{9-\sqrt{9^2-80}}{2}}\)=\(\sqrt{5}\)\(-\)\(\sqrt{4}\)=\(2-\sqrt{5}\)
(dựa theo công thức có sẵn từ một quyển sách nâng cao:\(\sqrt{A\pm\sqrt{B}}\)=\(\sqrt{\dfrac{A+\sqrt{A^2-B}}{2}}\pm\sqrt{\dfrac{A-\sqrt{A^2-B}}{2}}\)
c: \(\Leftrightarrow4x^2-6x+9=16\)
\(\Leftrightarrow4x^2-6x-7=0\)
hay \(x\in\left\{\dfrac{3+\sqrt{37}}{4};\dfrac{3-\sqrt{37}}{4}\right\}\)
d: \(=\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\)
\(=\dfrac{1}{2}\sqrt{3}+\dfrac{5}{2}\)
Giải phương trình:
1, \(x^2+2x\sqrt{x-\dfrac{1}{x}}=3x+1\)
2, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{16x-4x^2-15}\)
3, \(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
giải pt:
a) \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\)
b) \(3x+\sqrt{4x^2-8x+4}=1\)
c) \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)
giúp mk vs ạ mk cần gấp