Bài 4: Liên hệ giữa phép chia và phép khai phương

Câu 2: 

a: ĐKXĐ: x>=0; x<>1

b: \(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=-\dfrac{2\sqrt{x}}{2}\cdot\left(\sqrt{x}-1\right)=-\sqrt{x}\left(\sqrt{x}-1\right)\)

c: Thay x=4/25 vào G, ta được:

\(G=-\dfrac{2}{5}\cdot\left(\dfrac{2}{5}-1\right)=\dfrac{-2}{5}\cdot\dfrac{-3}{5}=\dfrac{6}{25}\)

a) ( \(\sqrt{12}\) + \(\sqrt{3}\) ) (\(\sqrt{27}\) - \(\sqrt{3}\) )

= (\(\sqrt{3}.\sqrt{4}\) + \(\sqrt{3}\) ) ( \(\sqrt{3}.\sqrt{9}\) - \(\sqrt{3}\) )

= \(\sqrt{3}\left(\sqrt{4}+1\right)\). \(\sqrt{3}\left(\sqrt{9}-1\right)\)

= \(\sqrt{3}.3.\sqrt{3}.2\)

= 3. 6 = 18

b) \(\left(5\sqrt{3}-2\sqrt{7}\right)\)\(\left(5\sqrt{3}+2\sqrt{7}\right)\)

=\(\left(5\sqrt{3}\right)^2\) \(-\left(2\sqrt{7}\right)^2\)

= 75 - 28 = 47

\(A=\sqrt{-x^2+x+\dfrac{3}{4}}=\sqrt{-\left[\left(x^2-x+\dfrac{1}{4}\right)-1\right]}=\sqrt{-\left[\left(x-\dfrac{1}{2}\right)^2-1\right]}\le\sqrt{1}=1\Leftrightarrow x=\dfrac{1}{2}\)

x = 0 , 5

a: \(=\sqrt{3a}:\sqrt{b}\)

b: \(=\sqrt{a}:\sqrt{xy}\)

1. c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)

\(\Leftrightarrow6+7\sqrt{6}-2\sqrt{6}-14\)

\(\Leftrightarrow-8+5\sqrt{6}\)

d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)

\(\Leftrightarrow3-5\sqrt{3}+2\sqrt{3}-3\)

\(\Leftrightarrow-3\sqrt{3}\)

@ngonhuminh @ngonhuminh

Bài 2: 

a: \(=\dfrac{4\sqrt{3}-3\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=4-3+8=9\)

b: \(=\dfrac{2\sqrt{5}-9\sqrt{5}+36\sqrt{5}}{\sqrt{5}}=2-9+36=29\)

c: \(=\dfrac{4\sqrt{5}-9\sqrt{5}+16\sqrt{5}}{\sqrt{5}}=4-9+16=11\)

d: \(=\dfrac{6\sqrt{6}+12\sqrt{6}-20\sqrt{6}}{\sqrt{6}}=18-20=-2\)

Câu 1: 

a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b: Để Q>0 thì \(\sqrt{a}-2>0\)

=>a>4

Lời giải:

a)

ĐK: \(\forall x\in\mathbb{R}\)

Ta có: \(\sqrt{3x^2}-\sqrt{12}=0\)

\(\Rightarrow \sqrt{3x^2}=\sqrt{12}\)

\(\Rightarrow 3x^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2\) (đều thỏa mãn)

b) ĐK: \(\forall x\in\mathbb{R}\)

\(\sqrt{(x-3)^2}=9\)

\(\Leftrightarrow |x-3|=9\Rightarrow \left[\begin{matrix} x-3=9\\ x-3=-9\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=12\\ x=-6\end{matrix}\right.\)

c) ĐK: $x\in\mathbb{R}$
\(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow \sqrt{(2x)^2+2.2x+1}=6\)

\(\Leftrightarrow \sqrt{(2x+1)^2}=6\)

\(\Leftrightarrow |2x+1|=6\)

\(\Rightarrow \left[\begin{matrix} 2x+1=6\\ 2x+1=-6\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=-\frac{7}{2}\end{matrix}\right.\)

d) ĐK: \(x\geq 1\)

\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)

\(\Leftrightarrow \sqrt{16(x-1)}-\sqrt{9(x-1)}+\sqrt{4(x-1)}+\sqrt{x-1}=8\)

\(\Leftrightarrow 4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)

\(\Leftrightarrow 4\sqrt{x-1}=8\Rightarrow \sqrt{x-1}=2\)

\(\Rightarrow x=2^2+1=5\) (thỏa mãn)

e)

ĐK: \(-4\leq x\leq \frac{1}{2}\)

\(\sqrt{1-x}+\sqrt{1-2x}=\sqrt{x+4}\)

\(\Leftrightarrow \sqrt{1-x}-1+\sqrt{1-2x}-1=\sqrt{x+4}-2\)

\(\Leftrightarrow \frac{(1-x)-1}{\sqrt{1-x}+1}+\frac{(1-2x)-1}{\sqrt{1-2x}+1}=\frac{(x+4)-2^2}{\sqrt{x+4}+2}\)

\(\Leftrightarrow \frac{-x}{\sqrt{1-x}+1}+\frac{-2x}{\sqrt{1-2x}+1}=\frac{x}{\sqrt{x+4}+2}\)

\(\Leftrightarrow x\left(\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{1-x}+1}+\frac{2}{\sqrt{1-2x}+1}\right)=0\)

Dễ thấy biểu thức trong ngoặc lớn lớn hơn $0$

Do đó: \(x=0\) là nghiệm duy nhất của pt.

Lời giải:

a) ĐKXĐ: \(a>0; a\neq 1\)

\(M=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\frac{a-1}{2\sqrt{a}}.\frac{(a-\sqrt{a})(\sqrt{a}-1)-(a+\sqrt{a})(\sqrt{a}+1)}{(\sqrt{a}+1)(\sqrt{a}-1)}\)

\(=\frac{a-1}{2\sqrt{a}}.\sqrt{a}.\frac{(\sqrt{a}-1)^2-(\sqrt{a}+1)^2}{a-1}\)

\(=\frac{(\sqrt{a}-1)^2-(\sqrt{a}+1)^2}{2}=\frac{a+1-2\sqrt{a}-(a+1+2\sqrt{a})}{2}=\frac{-4\sqrt{a}}{2}=-2\sqrt{a}\)

b)

Để \(M=-4\Leftrightarrow -2\sqrt{a}=-4\Leftrightarrow \sqrt{a}=2\Rightarrow a=4\)

\(\left(\sqrt{12}+\sqrt{3}\right)\left(12+3\right)\left(\sqrt{27}-\sqrt{3}\right)\left(27-3\right)\)

\(=3\sqrt{3}.15.2\sqrt{3}.24\)

= \(3\sqrt{3}.2\sqrt{3}.15.24\)

= \(18.15.24=6480\)

\(\left(5\sqrt{3}-2\sqrt{7}\right)\left(5\sqrt{3}+2\sqrt{7}\right)\left(53-27\right)\left(53+27\right)\)

= \(\left(\left(5\sqrt{3}\right)^2-\left(2\sqrt{7}\right)^2\right).26.80\)

= \(\left(75-28\right).26.80\)

= 47.26.80 = 97760