sao khó vậy,mình học lớp 9 mà tính mãi chẳng ra đáp án bài này từ lâu rùi

Bài 1 : 

\(2+\sqrt{9}=2+3=5\)

Bài 2 : 

Với \(x\ge0\)

\(B=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}+7}\right):\frac{5}{\sqrt{x}+7}\)

\(=\frac{\sqrt{x}+7-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}:\frac{5}{\sqrt{x}+7}\)

\(=\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}.\frac{\sqrt{x}+7}{5}=\frac{1}{\sqrt{x}+2}\)

Bài 3 : 

\(\hept{\begin{cases}x+2y=4\left(1\right)\\x-2y=0\left(2\right)\end{cases}}\)Lấy (1) - (2) ta được : 

\(4y=4\Leftrightarrow y=1\)

Thay y = 1 vào (1) ta được : \(x+2=4\Leftrightarrow x=2\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

1.

\(2 +\sqrt{9}=2+3=5\)

2.

\(B =\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}+7}\right):\frac{5}{\sqrt{x}+7}\)

\(B=\left[\frac{\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}\right]:\frac{5}{\sqrt{x}+7}\)

\(B=\left[\frac{\sqrt{x}+7-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}\right]:\frac{5}{\sqrt{x}+7}\)

\(B=\left[\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}\right].\frac{\sqrt{x}+7}{5}\)

\(B=\frac{1}{\sqrt{x}+2}\)

Vậy \(B=\frac{1}{\sqrt{x}+2}\)khi x \(\ge\)0

3. 

\(\hept{\begin{cases}x+2y=4\\x-2y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4-2y\\4-2y-2y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4-2y\\y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=1\end{cases}}}\)

Vậy HPT có 2 nghiệm duy nhất \(\left(x;y\right)=\left(1;2\right)\)

a,Ta có  \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)

b, Với \(x\ge0;x\ne1\)

 \(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)

\(=\left(\frac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\right)\)

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

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

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a. Ta có $x=4-2\sqrt{3}=1-\mathrm{2.1.}\sqrt{3}+3=(1-\sqrt{3}{)}^2$.

$\Rightarrow \sqrt{x}=|1-\sqrt{3}|=\sqrt{3}-1.$

1x−1=13−1−1=13−2=−2−3.

b. Với $x\ge 0$ và $x\ne 1$ ta có:

−3xxx−1−11−x):(1−x+21+x+x)

−3x(x−1)(x+x+1)+1x−1]:(1+x+x)−(x+2)1+x+x

−3x+x+x+1(x−1)(x+x+1).x+x+1x−1

x−2x+1(x−1)(x+x+1).x+x+1x−1 

a, Với \(x\ge0,x\ne4\)

\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

b, Ta có  \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)

\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)

1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

2 , A = \(1-\sqrt{2}\)

ĐKXĐ : \(y>-5\)

Đặt \(\left(x-2\right)^2=a>0\) và \(\frac{1}{\sqrt{y+5}=b}\)

Hệ phương trình đã cho trở thành : \(\hept{\begin{cases}2a+b=3\\a-2b=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}4a+2b=6\\a-2b=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5a=5\\a-2b=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=1\end{cases}}\)( Thỏa mãn )

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}=1}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\\\sqrt{y+5}=1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}=1}\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{y+5}=1\\\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\end{cases}\Leftrightarrow}\hept{\begin{cases}y+5=1\\\orbr{\begin{cases}x=3\\x=1\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x=3\\y=-4\end{cases}}\\\hept{\begin{cases}x=1\\y=-4\end{cases}}\end{cases}}}\)

ĐKXĐ : y > -5

Đặt \(\hept{\begin{cases}\left(x-2\right)^2=a\\\frac{1}{\sqrt{y+5}}=b\end{cases}\left(a\ge0;b>0\right)}\)

Hpt đã cho trở thành \(\hept{\begin{cases}2a+b=3\\a-2b=-1\end{cases}}\)=> \(a=b=1\left(tm\right)\)

=> \(\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}}=1\end{cases}}\)<=> \(\hept{\begin{cases}x=3\\y=-4\end{cases}}or\hept{\begin{cases}x=1\\y=-4\end{cases}}\)(tm)

Vậy ... 

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Ta có đẳng thức : (2−3+√3√3+1).(2+3−√3√3−1)=1

xét vế trái ta có :(2−3+√3√3+1).(2+3−√3√3−1)  = 

a) ta co \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)=1\)

b) ta co \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)

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

             \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)

             \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Vay \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)

=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)

=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)

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

=-4

2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)

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

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

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

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

1. (1−5+√2√2+1)⋅√3+2√2=−4√2+1√(√2+1)2=−4(1−5+22+1)⋅3+22=−42+1(2+1)2=−4.

2. Với x>0;x≠1x>0;x≠1 ta có:
A=(√xx+√x−1√x−1):2x+√x−2A=(xx+x−1x−1):2x+x−2
⇔A=(√x√x(√x+1)−1√x−1):2(√x−1)(√x+2)⇔A=(xx(x+1)−1x−1):2(x−1)(x+2)
⇔A=−2(√x−1)(√x+1)⋅(√x−1)(√x+2)2⇔A=−2(x−1)(x+1)⋅(x−1)(x+2)2
⇔A=−(√x+2)√x+1⇔A=−(x+2)x+1. Vạyy với x>0;x≠1x>0;x≠1, ta có A=−(√x+2)√x+1A=−(x+2)x+1.

Bài 1:

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

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

2) \(VT=\dfrac{\left(6a+1\right)\left(a+6\right)+\left(6a-1\right)\left(a-6\right)}{a\left(a-6\right)\left(a+6\right)}.\dfrac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\dfrac{12a^2+12}{a\left(a^2+1\right)}=\dfrac{12\left(a^2+1\right)}{a\left(a^2+1\right)}=\dfrac{12}{a}=VP\)

Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

1.

\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)

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

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

2.

\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)

\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

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