\(\sqrt[]{\dfrac{\text{1}}{\text{4}}}\).\(\sqrt[]{\text{1,21}}\)-\(\sqrt[]{\text{0,09}}\)+\(\sqrt[]{\text{36}}\)
Những câu hỏi liên quan
-5.\(\sqrt{\text{0,25}}\)+\(\sqrt{\text{195}}\)0-5.\(\sqrt{\dfrac{\text{36}}{\text{25}}}\)
\(=-5\cdot\dfrac{1}{2}+0-5\cdot\dfrac{6}{5}=-\dfrac{5}{2}-6=-\dfrac{17}{2}\)
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1, P=(\(\dfrac{\text{x-1}}{\text{x+3}\sqrt{\text{x-4}}}+\dfrac{\sqrt{\text{x}}+1}{1-\sqrt{\text{x}}}\)) : \(\dfrac{\text{x}+2\sqrt{\text{x}}+1}{x-1}\)+1
a, Rút gọn P
b, Tìm x để P<0
(\(\dfrac{\text{3}}{\text{2}}\).\(\sqrt[]{\dfrac{\text{4}}{\text{25}}+}\)3.\(\sqrt[]{\text{0,04}}\)):\(\sqrt[]{\dfrac{\text{9}}{\text{64}}}\)
\(=\left(\dfrac{3}{2}\cdot\dfrac{2}{5}+2\cdot\dfrac{1}{5}\right):\dfrac{3}{8}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}=\dfrac{8}{3}\)
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\(\text{}\text{}\text{}\text{}\dfrac{2\left(4-2\sqrt{3}\right)-3\sqrt{4-2\sqrt{3}}-2}{\sqrt{4-2\sqrt{3}}-2}\)
\(=\dfrac{8-4\sqrt{3}-3\left(\sqrt{3}-1\right)-2}{\sqrt{3}-1-2}=\dfrac{6-4\sqrt{3}-3\sqrt{3}+3}{\sqrt{3}-3}\)
\(=\dfrac{-7\sqrt{3}+3}{\sqrt{3}-3}=3\sqrt{3}+2\)
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1 a..Rút gọn biểu thức A = \(\dfrac{\text{ x 2 − 4 x + 4}}{\text{x 3 − 2 x 2 − ( 4 x − 8 ) }}\)
b. Rút gọn biểu thức B = \(\left(\dfrac{x+2}{\text{x }\sqrt{\text{x }}+1}-\dfrac{1}{\sqrt{\text{x}}+1}\right).\dfrac{\text{4 }\sqrt{x}}{3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
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\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
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Giaỉ phương trình:
a) \(\sqrt{16\text{x}-48}-6\sqrt{\dfrac{x-3}{4}}+\sqrt{4\text{x}-12}=5\)
b) \(\sqrt{1-10\text{x}+25\text{x}^2}-4=2\)
giải phương trình: \(\sqrt{\text{x}^2-\text{x}+1}+\sqrt{-2\text{x}^2+\text{x}+2}=\dfrac{\text{ }\text{x}^2-4\text{x}+7}{2}\)
Giải bằng bất đẳng thức Cô si: (ĐK: \(x^2-x+1\ge0;-2x^2+x+2\ge0;x^2-4x+7\)
Ta có: \(x^2-x+1+1\ge2\sqrt{x^2-x+1}\Leftrightarrow\sqrt{x^2-x+1}\le\dfrac{x^2-x+2}{2}\left(1\right)\\ T,T:\sqrt{-2x^2+x+2}\le\dfrac{-2x^2+x+3}{2}\left(2\right)\\ \left(1\right);\left(2\right)\Rightarrow\sqrt{x^2-x+1}+\sqrt{-2x^2+x+2}\le\dfrac{x^2-x+2-2x^2+x+3}{2}=\dfrac{-x^2+5}{2}\\ \Rightarrow\sqrt{x^2-x+1}+\sqrt{-2x^2+x+2}-\dfrac{x^2-4x+7}{2}\le\dfrac{-x^2+5-x^2+4x-7}{2}\\
=\dfrac{-2x^2+4x-2}{2}\\
=-x^2+2x-1
\\
\Rightarrow-\left(x-1\right)^2\ge0\)
Điều này chỉ thỏa 1 điều kiên khi x-1=0 ⇔x=1(nhận
Vậy x=1 là nghiệm cuả phương trình
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4.\(\sqrt{\text{25}}\)-2\(\sqrt[]{\dfrac{\text{4}}{\text{9}}}\)
\(=4\cdot5-2\cdot\dfrac{2}{3}=20-\dfrac{4}{3}=\dfrac{56}{3}\)
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\(\left(\dfrac{\text{√}x}{\text{√}x+2}+\dfrac{8\text{√}x+8}{x+2\text{√}x}-\dfrac{\text{√}x+2}{\text{√}x}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
a) rút gọn P
b)CMR: P≤1
b) (4√x + 4)/(x + 2√x + 5) ≥ 1
⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
Do x ≥ 0 ⇒ x + 2√x + 5 > 0
⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0
⇔ 4√x + 4 - x - 2√x - 5 ≤ 0
⇔ -x + 2√x - 1 ≤ 0
⇔ -(x - 2√x + 1) ≤ 0
⇔ -(√x - 1)² ≤ 0 (luôn đúng)
Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0
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a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+5}\)
b: 4(căn x+1)>=4
x+2căn x+5>=5
=>P<=4/5<1
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