Rút gọn :\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\left(a\ge0\right)\)
Rút gọn
\(2x-\sqrt{4x^2+4x+1}\)với x bé hơn -1/2
\(3x+2-\sqrt{9x^2-12x+4}\)
\(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}vớia\ge0\)
\(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}vớia\ge0\)
\(\frac{3}{a^2-b^2}\sqrt{\frac{2a^2-4ab+2b^2}{9}}\)
\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\frac{\sqrt{a}+1}{a}\)
a) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
Vì \(x< -\frac{1}{2}\)nên \(\left|2x+1\right|=-\left(2x+1\right)\)
\(\Rightarrow2x+2x+1=4x+1\)
b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)
Khi \(x\ge\frac{2}{3}\)thì \(\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)
Khi \(x< \frac{2}{3}\) thì \(\left|3x-2\right|=2-3x\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
Đặt \(\sqrt{a}=x\) ta được : \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do \(a\ge0\))
d) \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)
TK NKA !!!
Tính
a)\(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
b)\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) với \(a\ge0\)
c)\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)
d)\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right).\sqrt{11}+3\sqrt{32}\)
Lời giải:
a)
$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$
$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$
b)
$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$
$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$
$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$
c)
$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$
$=6-\sqrt{15}$
d)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$
Rút Gọn
A. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\)\(\sqrt{7}\)+\(7\sqrt{8}\)
B.\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
C.\(\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{15-6\sqrt{6}}\)
D.\(\left(\sqrt{3}\right)+1\sqrt{4-2\sqrt{2}}\)
E.2\(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
Khai triển và rút gọn các biểu thức sau
a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)( a lớn hơn hoặc bằng 0)
b) \(\left(4\sqrt{x}-\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)\)
c) \(\dfrac{2}{2x-1}\sqrt{5x^2\left(1-4x+4x^2\right)}\)
( x lớn hơn 0,5)
a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)
b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)
\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)
\(=\left(6-5\sqrt{2}\right)x\)
c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)
1) thực hiện phép tính và rút gọn biểu thức
a) A=\(\sqrt{9a}-\sqrt{16a}-\sqrt{49a}\) với \(a\ge0\)
b) B=\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
lm nhanh giúp mk nhé mk đang cần gấp
a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)
b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=2+\sqrt{3}+\sqrt{2}+1-\sqrt{3}-\sqrt{2}=3\)
Rút gọn
a)\(2\sqrt{a}+3a\sqrt{4ab^2}-2b\sqrt{16a^5}-2\sqrt{25a}\)(a>0;b>0)
b)\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\left(a\ge0;b\ge0;a\ne b\right)\)
c)\(\frac{a\sqrt{a}-b\sqrt{b}}{a-b}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0;a\ne0\right)\)
Rút gọn biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)1>Cho biểu thức: Q= \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)
a) Rút gọn biểu thức Q
b) tìm x để Q = \(\dfrac{6}{5}\)
c) tìm các giá trị nguyên của x để biểu thức Q có giá trị nguyên
2> Tính:
a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\left(a\ge0\right)\)
b)\(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\)
c)\(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
Các bạn giúp mình với .Cảm ơn!!
đk : \(x\ne4\) ; \(x\ge0\)
1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\) = \(\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{3}{2+\sqrt{x}}\)
b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)
\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)
c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4
ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên
\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3
mà \(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3
\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)
vậy x = 1 nguyên thì Q nguyên
2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)
= \(4\sqrt{a}-5\sqrt{10a}\)
b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)
c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-2\sqrt{22}-11+3\sqrt{22}\)
= \(22+\sqrt{22}\)
Rút gọn
\(a.\sqrt{9\text{8}}-\sqrt{72}\text{+}\frac{\text{1}}{2}\sqrt{\text{8}}\)
b.\(\sqrt{\text{16}a}\text{+}2\sqrt{40a}-\text{3}\sqrt{90a}\)
c.\(\left(2\sqrt{\text{3}}\text{+}\sqrt{\text{5 }}\right)\sqrt{\text{3}}-\sqrt{\text{6}0}\)
a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)
c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
Rút gọn
a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)
\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)
\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)
\(=\sqrt{2}\left(1+1\right)=2\sqrt{2}\)
b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)
\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)
\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)
\(=4\sqrt{a}-5\sqrt{10a}\)
c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-\sqrt{60}\)
\(=6-\sqrt{15}\)