Rút gọn
\(a.\sqrt{9\text{8}}-\sqrt{72}\text{+}\frac{\text{1}}{2}\sqrt{\text{8}}\)
b.\(\sqrt{\text{16}a}\text{+}2\sqrt{40a}-\text{3}\sqrt{90a}\)
c.\(\left(2\sqrt{\text{3}}\text{+}\sqrt{\text{5 }}\right)\sqrt{\text{3}}-\sqrt{\text{6}0}\)
Những câu hỏi liên quan
Rút gọn :
a.text{3}sqrt{2}text{+ }4sqrt{text{8}}-sqrt{text{1}text{8}}
b.sqrt{text{3}}-frac{text{1}}{text{3}}sqrt{27}text{+ }2sqrt{text{5}07}
c.sqrt{2text{5}a}text{+ }sqrt{49a}-sqrt{text{6}4a}
d.-sqrt{text{3}text{6}b}text{−}frac{text{1}}{text{3}}sqrt{text{5}4b}text{+}frac{text{1}}{text{5}}sqrt{text{1}text{5}0b}
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Rút gọn :
a.\(\text{3}\sqrt{2}\text{+ }4\sqrt{\text{8}}-\sqrt{\text{1}\text{8}}\)
b.\(\sqrt{\text{3}}-\frac{\text{1}}{\text{3}}\sqrt{27}\text{+ }2\sqrt{\text{5}07}\)
c.\(\sqrt{2\text{5}a}\text{+ }\sqrt{49a}-\sqrt{\text{6}4a}\)
d.\(-\sqrt{\text{3}\text{6}b}\text{−}\frac{\text{1}}{\text{3}}\sqrt{\text{5}4b}\text{+}\frac{\text{1}}{\text{5}}\sqrt{\text{1}\text{5}0b}\)
a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=\sqrt{2}\left(3+4\cdot2-3\right)\)
\(=8\sqrt{2}\)
b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)
\(=\sqrt{3}\left(1-1+26\right)\)
\(=26\sqrt{3}\)
c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)
\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)
\(=\sqrt{a}\left(5+7-8\right)\)
\(=4\sqrt{a}\)
d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)
\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)
\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)
\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)
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rút gọn các biểu thức sau
a) \(\frac{4}{\sqrt{10}}\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)\)
b)\(\left(4+\sqrt{\text{15}}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{\text{4}-\sqrt{15}}\)
c)\(\sqrt{\text{4 }\sqrt{\text{6}}\text{ }+8\sqrt{\text{3 }}+4\sqrt{2}+18}\)
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
Bài 1: cho P= \(\left(\frac{\sqrt{b}}{\sqrt{a}-\sqrt{a-b}}\text{+}\frac{\sqrt{b}}{\sqrt{a}\text{+}\sqrt{a\text{+}b}}\right)\div\left(1\text{+}\frac{\sqrt{a\text{+}b}}{\sqrt{a-b}}\right)\)
a) Rút gọn P
b) Tính P khi a=\(24-2\sqrt{15}\); b=16
Bài 2: \(x\ge0\), cm: \(x^2-3\sqrt{x}\text{+}\frac{5}{2}\)>0
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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\(\left(\dfrac{\text{√}x}{\text{√}x+2}+\dfrac{8\text{√}x+8}{x+2\text{√}x}-\dfrac{\text{√}x+2}{\text{√}x}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
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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Bài 1 Rút gọn và tínha, sqrt{frac{sqrt{a}-1}{sqrt{b}+1}}:sqrt{frac{sqrt{b}-1}{sqrt{a}+1}}với a7.25;b3.25b,sqrt{15text{a}^2-8text{a}sqrt{15}+16} với asqrt{frac{3}{5}}+sqrt{frac{5}{3}}c,sqrt{10text{a}^2-4text{a}sqrt{10}+4}với asqrt{frac{2}{5}}+sqrt{frac{5}{2}}d,sqrt{a^2+2text{a}sqrt{a^2-1}}-sqrt{a^2-2sqrt{a^2-1}}vtext{ới}asqrt{5}
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Bài 1 Rút gọn và tính
a, \(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\)với a=7.25;b=3.25
b,\(\sqrt{15\text{a}^2-8\text{a}\sqrt{15}+16}\) với \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\)
c,\(\sqrt{10\text{a}^2-4\text{a}\sqrt{10}+4}\)với \(a=\sqrt{\frac{2}{5}}+\sqrt{\frac{5}{2}}\)
d,\(\sqrt{a^2+2\text{a}\sqrt{a^2-1}}-\sqrt{a^2-2\sqrt{a^2-1}}v\text{ới}\)\(a=\sqrt{5}\)
Bài 1: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
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Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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Rút gọn các biểu thức sau: 9, A sqrt{4+sqrt{15}}-sqrt{7-3sqrt{5}}10, A sqrt{2+sqrt{3}}+sqrt{2-sqrt{3}}11, A text{}text{}text{}sqrt{12-3sqrt{7}}-sqrt{12+3sqrt{7}}12, A left(3sqrt{2}+sqrt{6}right)sqrt{6-3sqrt{3}}13, A sqrt{9-4sqrt{5}}-sqrt{14-6sqrt{5}}
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Rút gọn các biểu thức sau:
9, A = \(\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
10, A = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
11, A = \(\text{}\text{}\text{}\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
12, A = \(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}\)
13, A = \(\sqrt{9-4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
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\(choP=\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)a;R\text{ú}tg\text{ọ}nP....b;T\text{í}nhPkhiX=3-2\sqrt{2}c;t\text{ì}mX\text{đ}\text{ể}P=1\)