Cho biểu thức Q = t ( 2 t 3 + t + 2 ) - 2 t 2 ( t 2 + 1 ) + t 2 - 2 t + 1 . Chứng tỏ giá trị của Q không phụ thuộc vào giá trị của t.
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Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
1: Ta có: \(\sqrt{x^2-x+\frac{1}{4}}\)
\(=\sqrt{x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2}\)
\(=\sqrt{\left(x-\frac{1}{2}\right)^2}\)
\(=\left|x-\frac{1}{2}\right|\)
2: Ta có: \(\sqrt{x^2}+\sqrt{x^6}\)
\(=\sqrt{x^2}\cdot1+\sqrt{x^2}\cdot\sqrt{x^4}\)
\(=\sqrt{x^2}\cdot\left(1+\sqrt{x^4}\right)\)
\(=\left|x\right|\cdot\left(1+x^2\right)\)
3: Ta có: \(C=\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot1+1}-\sqrt{4-2\cdot2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}-1\right|-\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}-1-2+\sqrt{2}\)
\(=2\sqrt{2}-3\)
bài 1: a) \(A=\frac{\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right)}{\frac{a+2}{a-2}}\)
\(A=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{a+2}{a-2}\)
\(A=\left(\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\right)\cdot\frac{a-2}{a+2}\)
\(A=2\cdot\frac{a-2}{a+2}\left(a\ne0;a\ne\pm2\right)\)
b) để A = 1 => \(2\cdot\frac{a-2}{a+2}=1\)
=> 2a - 4 = a + 2
=> a = 6 (thỏa mãn)
bài 2) a) ĐKXĐ: \(x\ne4\)
b) \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(\Leftrightarrow B=\frac{2\sqrt{x}+\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow B=\frac{2\sqrt{x}+4}{x-4}=\frac{2}{\sqrt{x}-2}\)
c) \(B=\frac{2}{\sqrt{3+2\sqrt{3}}-2}\) \(\approx3,69\)
(bạn tự bấm máy tính nhé nhưng theo mình thấy nếu x = 4 + 2\(\sqrt{3}\) hay \(3+2\sqrt{2}\) thì sẽ cho kết quả đẹp hơn, k biết bạn có nhầm đề k nữa!)
a)đkxđ:\(x\ge0,x\ne4\)
P=\(\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x+2}}\)
\(=\left(\dfrac{2}{\left(\sqrt{x-2}\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\sqrt{x}+2\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Rút gọn được Q = 1 Þ đpcm.