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a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)
b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
nhờ bạn làm rõ vì sao \(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2-\sqrt{x}}{x-1}\) lại bằng \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
mình xin cảm ơn
1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
\(\frac{\sqrt{x^2}+\sqrt{4-4x+x^2+1}}{2x-1}\)
\(=\frac{x+2-2\sqrt{x}+1}{2x-1}\)
\(=1+\frac{4-2\sqrt{x}}{2x-1}\)
em lớp 8 chỉ làm được thế thôi
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)
\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)
\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)
\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)
\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)
\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)
\(=4\sqrt{x}\cdot\left(-1\right)\)
\(=-4\sqrt{x}\)
a.
\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)
\(B=\sqrt{16}.\sqrt{x+1}-\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}\)
\(B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(B=\left(4-3+2+1\right).\sqrt{x+1}\)
\(B=4.\sqrt{x+1}\)
b.
\(B=16\\\)
\(\Rightarrow4\sqrt{x+1}=16\)
\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)
\(\Rightarrow x+1=4^2\)
\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)
vậy x=15
Bài làm :
1) Khi x=9 ; giá trị của A là :
\(A=\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)
2) Ta có :
\(B=...\)
\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
3) Ta có :
\(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+2}\div\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)
Xét :
\(\frac{A}{B}+1=\frac{4}{\sqrt{x+2}}>0\Rightarrow\frac{A}{B}>-1\)
=> Điều phải chứng minh
1, thay x=9(TMĐKXĐ) vào A ta đk:
A=\(\dfrac{\sqrt{9}}{\sqrt{9}-2}=3\)
vậy khi x=9 thì A =3
2,với x>0,x≠4 ta đk:
B=\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
3,\(\dfrac{A}{B}>-1\) (x>0,x≠4)
⇒\(\dfrac{\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}>-1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}>-1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>-1\)
⇒\(\sqrt{x}-2>-1\) (vì \(\sqrt{x}+2>0\))
⇔\(\sqrt{x}>1\)⇔x=1 (TM)
vậy x=1 thì \(\dfrac{A}{B}>-1\) với x>0 và x≠4
a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
12b:
\(\left(\dfrac{1}{1-2\sqrt{x}}-1\right):\left(\dfrac{1}{4x-1}+\dfrac{1}{2\sqrt{x}+1}\right)\)
\(=\left(\dfrac{-1}{2\sqrt{x}-1}-1\right):\dfrac{1+2\sqrt{x}-1}{4x-1}\)
\(=\dfrac{-1-2\sqrt{x}+1}{2\sqrt{x}-1}\cdot\dfrac{4x-1}{2\sqrt{x}}\)
\(=\dfrac{-\left(4x-1\right)}{2\sqrt{x}-1}=-2\sqrt{x}-1\)