Rút gọn các biểu thức sau a)√27-✓12+✓48-5✓3 b)5✓18-✓5+✓20+✓1 2 C)✓25:✓16=✓36:✓9 D)✓12+✓27-5✓3 E)2✓3-✓75+2✓12
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
Rút gọn các biểu thức sau:
a) $E=2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt{\sqrt{75}}-4 \sqrt{15 \sqrt{27}}$ :
b) $F=\dfrac{1}{\sqrt{3}}+\dfrac{1}{3 \sqrt{2}}+\dfrac{1}{\sqrt{3}} \sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}} .$
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
rút gọn biểu thức chứa căn số học
a)-√20+3√45-6√80-1/5√125
b)2√3-√75+2√12-√147
c)3/2√12+7/5√75-9/10√300+11/6√108
\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)
\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)
\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)
a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)
\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)
\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)
b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)
\(=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\)
\(=-6\sqrt{3}\)
rút gọn biểu thức chứa căn số học
√27-2√3+2√48-3√75
\(=3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-6\sqrt{3}\)
Ta có: \(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\)
\(=3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-6\sqrt{3}\)
bài 1
a) rút gọn biểu thức : A= 4√12+ 3√75 -5√48
b) giải phương trình :√x-2 -√9x-18 =16
a ⇒A=\(4\sqrt{4\times3}+3\sqrt{25\times3}-5\sqrt{16\times3}=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}=3\sqrt{3}\)
b ĐKXĐ x≥2 ⇔\(\sqrt{x-2}+3\sqrt{x-2}=16\Leftrightarrow4\sqrt{x-2}=16\Leftrightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Leftrightarrow x=18\)
a. \(A=4\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)
\(=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)
\(=3\sqrt{3}\)
b. \(\sqrt{x-2}-\sqrt{9x-18}=16\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{9\left(x-2\right)}=16\)
\(\Leftrightarrow\sqrt{x-2}-3\sqrt{x-2}=16\)
\(\Leftrightarrow-2\sqrt{x-2}=16\)
\(\Leftrightarrow\sqrt{x-2}=-8\) ( Vô lý )
Vậy PT vô nghiệm
a, 4\(\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)
=\(8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)
= (8+15-20)\(\sqrt{3}\) = 3\(\sqrt{3}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}\)
b) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
c) \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}\)
d) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e) \(\dfrac{\sqrt{a}-\sqrt{b}^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\) với a > 0, b > 0
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
Rút gọn phân số :
a) 4/6 ; 12/8 ; 15/25 ; 11/22 ; 36/10 ; 75/36
b) 5/10 ; 12/36 ; 9/72 ; 75/300 ; 15/35 ; 4/100
Mẫu : 9/27 = 9:9/27:9 = 1/3
Tham khảo :
(Tại mik lười ko mún ghi nên bạn tham khảo cái ảnh này nhá)
Rút gọn các biểu thức sau:
$\sqrt[3]{0,001 x^{3}}, \quad \sqrt[3]{-125 a^{12}}, \quad \sqrt[3]{27 x^{6}}, \quad \sqrt[3]{-0,343 a^{3}}$
30,001x3=3(0,1x)3=0,1x;
\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};3−125a12=3(−5a4)3=−5a4;
\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};327x6=3(3x2)3=3x2;
\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;3−0,343a3=3(−0,7a)3=−0,7a;
Ta rút gọn các biểu thức như sau:
\(\sqrt[3]{0,001x^3}=\sqrt[3]{\left(0,1x\right)^3}=0,1x.\)
\(\sqrt[3]{-125a^{12}}=\sqrt[3]{\left(-5a^4\right)^3}=-5a^4\)
\(\sqrt[3]{27x^6}=\sqrt[3]{\left(3x^2\right)^3}=3x^2\)
\(\sqrt[3]{-0,343a^3}=\sqrt[3]{\left(-0,7a\right)^3}=-0,7a\)
\(\sqrt[3]{0,001x^3}=0,1x\) , \(\sqrt[3]{-125a^{12}}=-5a^4\) , \(\sqrt[3]{27x^6}=3x^2\) , \(\sqrt[3]{-0,343a^3}=-0,7a\)
rút gọn biểu thức
A=\(\sqrt{27}\)-2\(\sqrt{12}\)-\(\sqrt{75}\)
Ta có: \(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)
\(=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}\)
\(=-6\sqrt{3}\)
\(\Rightarrow A=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}=-4\sqrt{3}\)
⇔ A = \(3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\)
⇔ \(A=-6\sqrt{3}\)
