A = \(10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
Cho các biểu thức
a) rút gọn A và B
b) tìm x để A . B = -1
B = \(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}+\frac{2\sqrt{x}-1}{4-x}\)
A = \(10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
ĐKXĐ: \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2+\sqrt{x}-2-2\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{x-4}\)
\(A=10-\left(2\sqrt{2}-3\sqrt{3}\right)\left(-2\sqrt{2}-3\sqrt{3}\right)\)
\(=10+\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)\)
\(=10+\left(8-27\right)=-9\)
\(AB=-1\Leftrightarrow\frac{-9}{x-4}=-1\Rightarrow x-4=9\Rightarrow x=13\)
Tính:
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\\ =\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\\ =\left(-\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=9.3-2=25\)
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\sqrt{2}^2\)
\(=9.3-2=27-2=25\)
(\(\sqrt{32}-\sqrt{50}+\sqrt{27}\))(\(\sqrt{27}+\sqrt{50}-\sqrt{32}\))
= \(\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\)\(\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
= \(\left(3\sqrt{3}-\sqrt{2}\right)\)\(\left(3\sqrt{3}+\sqrt{2}\right)\)
= \(\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
= 27 - 2 = 25
Tính :
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)\)
\(=27-2\)
\(=25\)
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{3}-5\sqrt{2}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{2}-5\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+2\right)\)
\(\Leftrightarrow\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(\Rightarrow25\)
Vậy: BT = 25
P/s: từ dòng thứ 2 trở xuống bạn tự phân ... Vấn đề là ở bạn thôi :)))
\(=\left(\sqrt{27}\right)^2-\left(\sqrt{50}-\sqrt{32}\right)^2\)
\(=27-2=25\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
Rút gọn :
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}\)
2, \(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\)
3 , \(3\sqrt{112}-7\sqrt{216}+4\sqrt{54}-2\sqrt{252}-3\sqrt{96}\)
4, \(3\sqrt{3}\left(3+2\sqrt{6}-\sqrt{33}\right)\)
6, \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
7, \((5\sqrt{6}-4\sqrt{10}+7\sqrt{30}):\sqrt{2}\)
8, \(\left(2\sqrt{28}-3\sqrt{7}+5\sqrt{63}\right)\sqrt{112}\)
9, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
10, \(\left(4\sqrt{27}-2\sqrt{48}-5\sqrt{75}\right):2\sqrt{3}\)
11, \(\left(1+\sqrt{3}-\sqrt{2}\right).\left(1+\sqrt{3}+\sqrt{2}\right)\)
các bạn ơi ! giúp mik với đi !
Bài 1: Tính và rút gọn biểu thức:
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)\)
\(B=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(C=1-\left(\sqrt{45}-\sqrt{20}-\sqrt{3}\right)\left(\sqrt{20}-\sqrt{45}-\sqrt{3}\right)\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right):\frac{1}{\sqrt{6}}\)
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
tính
1/\(2\sqrt{20}-\sqrt{50}+3\sqrt{80}\)\(-\sqrt{320}\)
2/\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(1,4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)
\(2,\left(\sqrt{27}+\sqrt{32}-\sqrt{50}\right)\left(\sqrt{27}-\sqrt{32}+\sqrt{50}\right)\)
\(=27-\left(\sqrt{32}-\sqrt{50}\right)^2=27-2=25\)
Rút gọn các biểu thức:
1. A=\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
2. B= \(\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
3. C= \(\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)\)
4. D= \(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
5. E= \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
6. F= \(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)