Thực hiện các phép tính sau :
a. \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
b. \(\left(a^2b-3ab^2\right):\left(\dfrac{1}{2}ab\right)+\left(6b^3-5ab^2\right):b^2\)
K
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Những câu hỏi liên quan
25 tháng 11 2021
\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)
Y
8 tháng 7 2023
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
18 tháng 11 2021
\(a,=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\\ b,=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{x+1}{x+3}\left(x\ne-1;x\ne-2;x\ne-3\right)\\ =\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
18 tháng 11 2021
\(a,\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
\(b,\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)
\(=\left(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\left(\dfrac{x^2+4x+3}{\left(x+2\right)\left(x+3\right)}+\dfrac{x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{2x^2+8x+7}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{\left(2x^2+8x+7\right).x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+8x^2+7x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)
\(=\dfrac{2x^3+10x^2+15x+7}{\left(x+2\right)\left(x+3\right)^2}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right]:\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\left(a-\sqrt{ab}+\sqrt{b}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(B=\dfrac{a-\sqrt{ab}+b}{a-b}+\dfrac{2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{a-b}\)
\(B=\dfrac{a+\sqrt{ab}-b}{a-b}\)
G
8 tháng 8 2023
a) \(\sqrt{2}A=\sqrt{2x-2\sqrt{x-2}.\sqrt{x+2}}+\sqrt{2x+2\sqrt{x-2}.\sqrt{x+2}}\) (\(x\ge2\) )
\(=\sqrt{\left(x+2\right)-2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}+\sqrt{\left(x+2\right)+2\sqrt{x+2}.\sqrt{x-2}+\left(x-2\right)}\)
\(=\sqrt{\left(\sqrt{x+2}-\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}\)
\(=\left|\sqrt{x+2}-\sqrt{x-2}\right|+\sqrt{x+2}+\sqrt{x-2}\)
\(=\sqrt{x+2}-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\) ( do \(x+2>x-2\ge0\Leftrightarrow\sqrt{x+2}>\sqrt{x-2}\) )
\(=2\sqrt{x+2}\)
\(\Leftrightarrow A=\sqrt{2}.\sqrt{x+2}\)
Vậy...
b) \(B=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}.\dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{a-\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-b}{a-b}\)
Vậy...
9 tháng 11 2023
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
H
24 tháng 12 2023
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
24 tháng 12 2023
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
