a,\(\sqrt{\sqrt{\left(\sqrt{3-1}\right)^4}}\)
b,\(\sqrt{\sqrt{x^4}}\)vs \(x\ge0\)
M.N HELP T VS~~~ Ô - MAI - GÓT
Những câu hỏi liên quan
a)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright)left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
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a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
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Các b ơi giúp m vs
Câu 1: A = \(\frac{1}{2\sqrt{x}}+\frac{1}{2-\sqrt{x}}-\frac{2\sqrt{x}}{4-x}\left(x\ne4,x\ge0\right)\)0 và B = \(\left(\sqrt{2}+\sqrt{3}\right)\sqrt{2}-\sqrt{6}+\frac{\sqrt{333}}{\sqrt{111}}\)
a. Rút gọn A và B
b. Tìm x để A = B
a=căn (x)
A=[(4-a^2)(2-a)+2a(4-a^2)-4a^2(2-a)]/[(4-a^2)(2-a)2a]
A=(8-10a^2+4a+3a^3)/a(16-4a^2-8a+2a^3)
A=(a-2)^2(3a+2)/a(a+2)(a-2)^2*2
A=(3a+2)/a(a+2)*2
B=2+căn(3)
A=B suy ra
(3a+2)/a(a+2)*2=2+căn 3
<=>bấm máy tính ra nghiệm a=0.1539181357
=>x=a^2 =0.02341454985
tl đúng
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\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)ĐK:x\ge0.\)
Giúp mk bài này vs làm ko ra
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)
\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)
\(=x-2\sqrt{x}+4\)
=.= hok tốt!!
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Rút gọna)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright).left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
Đọc tiếp
Rút gọn
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right).\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
Giải giúp mình vs
1/ Thực hiện phép tính
a)sqrt{9-2sqrt{20}}+sqrt{12-2sqrt{35}}
b)sqrt{5-sqrt{21}}-sqrt{5+sqrt{21}}
2/ Rút gọn biểu thức
a) sqrt{dfrac{left(x-2right)^4}{left(3-xright)^2}}+dfrac{x^2-1}{x-3} (x 3)
b) 4x-sqrt{8}+dfrac{sqrt{x^3+2x^2}}{sqrt{x+2}} (x -2)
3/ Phân tích thành nhân tử
a) sqrt{x}+sqrt{y}-5sqrt{xy}-5yleft(x,yge0right)
b) xsqrt{x}-ysqrt{y}left(x,yge0right)
Đọc tiếp
Giải giúp mình vs ><
1/ Thực hiện phép tính
a)\(\sqrt{9-2\sqrt{20}}+\sqrt{12-2\sqrt{35}}\)
b)\(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
2/ Rút gọn biểu thức
a) \(\sqrt{\dfrac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\) (x < 3)
b) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\) (x > -2)
3/ Phân tích thành nhân tử
a) \(\sqrt{x}+\sqrt{y}-5\sqrt{xy}-5y\left(x,y\ge0\right)\)
b) \(x\sqrt{x}-y\sqrt{y}\left(x,y\ge0\right)\)
Bài 1 : Thực hiện phép tính :
a ) \(\sqrt{9-2\sqrt{20}}+\sqrt{12-2\sqrt{35}}\)
\(=\sqrt{5-2\sqrt{20}+4}+\sqrt{7-2\sqrt{35}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{5}\right)}^2\)
\(=\sqrt{5}-2+\sqrt{7}-\sqrt{5}\)
\(=\sqrt{7}-2\)
b ) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
\(=\sqrt{\dfrac{2\left(5-\sqrt{21}\right)}{2}}-\sqrt{\dfrac{2\left(5+\sqrt{21}\right)}{2}}\)
\(=\sqrt{\dfrac{10-2\sqrt{21}}{2}}-\sqrt{\dfrac{10+2\sqrt{21}}{2}}\)
\(=\dfrac{\sqrt{7-2\sqrt{21}+3}}{\sqrt{2}}-\dfrac{\sqrt{7+2\sqrt{21}+3}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
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Bài 3:
a: \(=\left(\sqrt{x}+\sqrt{y}\right)-5\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(1-5\sqrt{y}\right)\)
b: \(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)
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Giúp mình với:
Cho b/t A=\(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{x-9}\) (\(x\ge0\) ; \(x\ne9\) )
Rút gọn b/t A rồi tính giá trị tại x= \(2\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Mai Kt rồi :( :( :(
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
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giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
Dạng 1: Tính giá trị biểu thức [Rút gọn biểu thức rồi thay số (nếu đc)]
1) Tính giá trị biểu thức B sqrt{x-1+2sqrt[3]{xsqrt{x}+3x+3sqrt{x}+1}}, vs x 5
2) Tính giá trị biểu thức C sqrt{2x-1+2sqrt{x^2-x}+sqrt{2x-1-2sqrt{x^2-x}}}, vs x 4
3) Tính giá trị biểu thức D frac{sqrt[3]{xsqrt{x}left(3x+1right)+x^2left(3+xright)}}{sqrt{x}+1}-sqrt{x}, vs x 10
4) Tính giá trị biểu thức E sqrt{sqrt[4]{x}+1-2sqrt[8]{x}+1}, vs x 256
5) Cho x frac{left(sqrt{5}+2right)sqrt{3sqrt{5}-6}}{sqrt{4+sqrt{9-4...
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Dạng 1: Tính giá trị biểu thức [Rút gọn biểu thức rồi thay số (nếu đc)]
1) Tính giá trị biểu thức B = \(\sqrt{x-1+2\sqrt[3]{x\sqrt{x}+3x+3\sqrt{x}+1}}\), vs x = 5
2) Tính giá trị biểu thức C = \(\sqrt{2x-1+2\sqrt{x^2-x}+\sqrt{2x-1-2\sqrt{x^2-x}}}\), vs x = 4
3) Tính giá trị biểu thức D = \(\frac{\sqrt[3]{x\sqrt{x}\left(3x+1\right)+x^2\left(3+x\right)}}{\sqrt{x}+1}-\sqrt{x}\), vs x = 10
4) Tính giá trị biểu thức E = \(\sqrt{\sqrt[4]{x}+1-2\sqrt[8]{x}+1}\), vs x = 256
5) Cho x = \(\frac{\left(\sqrt{5}+2\right)\sqrt{3\sqrt{5}-6}}{\sqrt{4+\sqrt{9-4\sqrt{5}}}}\), tính giá trị biểu thức A = \(\left(x^4-5x^2+5\right)^{2014}\)
1) \(B=\sqrt{x-1+2\sqrt[3]{x\sqrt{x}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\sqrt{x^3}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\left(\sqrt{x}+1\right)^3}}\)
\(B=\sqrt{x-1+2\left(\sqrt{x}+1\right)}\)
\(B=\sqrt{x-1+2\sqrt{x}+2}\)
\(B=\sqrt{\left(\sqrt{x}+1\right)^2}\)
\(B=\sqrt{x}+1\)
\(B=\sqrt{5}+1\)
2) Sửa đề :
\(C=\sqrt{2x-1+2\sqrt{x^2-x}}+\sqrt{2x-1-2\sqrt{x^2-x}}\)
\(C=\sqrt{x+2\sqrt{x\left(x-1\right)}+x-1}+\sqrt{x-2\sqrt{x\left(x-1\right)}+x-1}\)
\(C=\sqrt{\left(\sqrt{x}+\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x}-\sqrt{x-1}\right)^2}\)
\(C=\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\)
\(C=2\sqrt{x}\)
\(C=2\cdot\sqrt{4}=4\)
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5) \(x=\frac{\left(\sqrt{5}+2\right)\sqrt{3\sqrt{5}-6}}{\sqrt{4+\sqrt{9-4\sqrt{5}}}}\)
\(\Leftrightarrow x=\frac{\left(\sqrt{5}+2\right)\cdot\sqrt{3}\cdot\sqrt{\sqrt{5}-2}}{\sqrt{4+\sqrt{\left(\sqrt{5}-2\right)^2}}}\)
\(\Leftrightarrow x=\frac{\left(\sqrt{5}+2\right)\cdot\sqrt{3}\cdot\sqrt{\sqrt{5}-2}}{\sqrt{4+\sqrt{5}-2}}\)
\(\Leftrightarrow x=\frac{\left(\sqrt{5}+2\right)\cdot\sqrt{3}\cdot\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+2}}\)
\(\Leftrightarrow x=\sqrt{3}\cdot\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\)
\(\Leftrightarrow x=\sqrt{3}\cdot\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(\Leftrightarrow x=\sqrt{3}\cdot\sqrt{5-4}\)
\(\Leftrightarrow x=\sqrt{3}\)
Thay vào A ta được :
\(A=\left[\left(\sqrt{3}\right)^4-5\cdot\left(\sqrt{3}\right)^2+5\right]^{2014}\)
\(A=\left(9-5\cdot3+5\right)^{2014}\)
\(A=\left(-1\right)^{2014}\)
\(A=1\)
Vậy...
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Xem thêm câu trả lời
cho biểu thức Q=\(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)với x\(\ge0,x\ne1\)
a) Rút gọn Q
b) tìm x để Q=3
Help me! Mai tớ phải nộp rồi ạ
a) Q=\(\left(\dfrac{2x+1}{\sqrt{x}^3-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
=\(\left(\dfrac{2x+1-x+\sqrt{x}}{\sqrt{x}^3-1}\right)\left(\dfrac{1+\sqrt{x}^3-\sqrt{x}-x}{1+\sqrt{x}}\right)\)
=\(\dfrac{\sqrt{x}+x+1}{\sqrt{x}^3-1}.\left(-2\sqrt{x}+1\right)\)
=\(\dfrac{\left(-2\sqrt{x}+1\right)\left(\sqrt{x}+x+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)=\(\dfrac{\left(-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
b) ta có : Q=3 => \(\dfrac{-2\sqrt{x}+1}{\sqrt{x}-1}=3=>-2\sqrt{x}+1=3\sqrt{x}-3\)
=>x=16/25=0,64
vậy x=0,64 khi Q=3
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a) Q=\(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
=\(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)-\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
=\(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)}{1+\sqrt{x}}\)
=\(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(x-2\sqrt{x}+1\right)\)
=\(\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\sqrt{x}-1\)
b) ta có : Q=3 <=> \(\sqrt{x}-1=3\)
\(\Leftrightarrow\) \(\sqrt{x}=4\Leftrightarrow x=16\)
vậy để Q=3 thì x=16
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