\(\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
b1: rút gọn biểu thức:
\(A=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7.\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
b2: tìm x, y, z thỏa mãn:
\(\sqrt{\left(x-\sqrt{2}\right)^2}_{ }\)+ \(\sqrt{\left(y+\sqrt{2}\right)^2}^{ }\)+ |x+y+z| = 0
nhanh nhé, ai đúng mk t*** cho !!!
1) Rút gọn biểu thức theo là cách hợp lý:
A = \(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
2) Tính hợp lý:
M = \(1-\frac{5}{\sqrt{196}}-\frac{5}{\left(2\sqrt{21}\right)^2}-\frac{\sqrt{25}}{204}-\frac{\left(\sqrt{5}\right)^2}{374}\)
3) Có hay không giá trị của x thỏa mãn điều kiện sau:
\(2002.\sqrt{\left(1+x\right)^2}+2003.\sqrt{\left(1-x\right)^2}=0\)
4) Tìm các số x, y, z thỏa mãn đẳng thức:
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
1A)thực hiện phép tính
a)\(\sqrt{144}.\sqrt{-\frac{-49}{64}}.\sqrt{0,01}\)
b)\(\left(\sqrt{0,25}-\sqrt{\left(-15\right)^2}+\sqrt{2,25}\right):\sqrt{169}\)
1b)hãy tính
a)\(\left(\sqrt{0,04}-\sqrt{\left(-1,2\right)^2}+\sqrt{121}\right).\sqrt{81}\)
b)\(75:\sqrt{3^2+\left(-4\right)^2}-3.\sqrt{\left(-5\right)^2-3^2}\)
B2 : Tính :
a, \(\left(\sqrt{x}-3\right)\)\(.\left(\sqrt{x}+2\right)\)
b, \(\left(\sqrt{x}-\sqrt{y}\right).\)\(\left(\sqrt{x}+\sqrt{y}\right)\)
c, \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right)\)\(.\sqrt{3}\)
d,\(\left(1+\sqrt{3}-\sqrt{5}\right)\)\(.\left(1+\sqrt{3}+\sqrt{5}\right)\)
a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)
b. \(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=x-y\)
c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)
\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{3}\right)^2-5=1+2\sqrt{3}+3-5=2\sqrt{3}-1\)
Tính
\(\dfrac{1}{x-y}\cdot\sqrt{x^4\left(x-y\right)^2}\) (x>y)
\(\sqrt{27}\cdot\sqrt{48\cdot\left(2-a\right)^2}\) (a>2)
\(\left(\sqrt{2012}+\sqrt{2011}\right)\cdot\left(\sqrt{2012}+\sqrt{2011}\right)\)
\(\sqrt{\dfrac{64x^2}{49\left(y+1\right)^2}}\) (x<0;y>-1)
\(\sqrt{\dfrac{121x^2}{144\left(y+2\right)}}\left(x>0;y< -2\right)\)
\(\sqrt{\dfrac{676x^3}{169xy^2}}\left(x>0;y< 1\right)\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
Chứng minh (với những giá trị của biến làm cho biểu thức có nghĩa)
a) \(\dfrac{\left(3\sqrt{xy}-6y-2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\dfrac{2\sqrt{xy}}{x-y}\right)=\sqrt{x}+\sqrt{y}\)
So sánh:
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\) với \(B=\sqrt{5}-\dfrac{3}{2}\)
Giúp với mình sắp cần rồi
Rút gọn
a.\(\left(2\sqrt{x}+\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)\)
b. \(\left(\sqrt{3x}+\sqrt{2x}\right)\left(3\sqrt{x}-\sqrt{6x}\right)\)
c.\(\left(\frac{4}{3}\sqrt{3}+\sqrt{2}\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{3}}\right)-2\)
d.\(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)(x,y lớn hơn hoặc bằng 0)
e.\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{x}\sqrt{y}+\sqrt{y}\right)\) (x,y lớn hơn hoặc bằng 0)