Rút gọn \(a+2b-\sqrt{a^2}-4ab+4b^2\)
bà 1 rút gọn biểu thức :\(\sqrt{9ab}\) + 7\(\sqrt{\dfrac{a}{b}}\) - 5\(\sqrt{\dfrac{b}{a}}\) - 3ab \(\sqrt{\dfrac{1}{ab}}\)
bài 2 :cho a>0,b>0 chứng minh : \(\dfrac{a^2b}{a-b}\).\(\sqrt{\dfrac{8\left(a^2-2ab+b^2\right)}{75a^4b}}\) = \(\dfrac{2}{15}\) .\(\sqrt{6b}\)
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
Rút gọn :
a) a2 + 4ab + 3b2 - 2b - 1
b) a2 - 2ab - 2b - 1
a, \(a^2+4ab+3b^2-2b-1=\left(a^2+4ab+4b^2\right)-\left(b^2+2b+1\right)=\left(a+2b\right)^2-\left(b+1\right)^2\)
\(=\left(a+2b-b-1\right)\left(a+2b+b+1\right)=\left(a+b-1\right)\left(a+3b+1\right)\)
b,\(a^2-2ab-2b-1=\left(a^2-2ab+b^2\right)-\left(b^2+2b+1\right)\)
\(=\left(a-b\right)^2-\left(b+1\right)^2\)
\(=\left(a-b-b-1\right)\left(a-b+b+1\right)\)
\(=\left(a-2b-1\right)\left(a+1\right)\)
TK MINK NHA!
a2 - 2ab - 2b - 1
= a2 - 2ab + b2 - b2 - 2b - 1
=( a - b )2 - ( b - 1 )2
= ( a - b - b + 1 ) ( a - b + b - 1 )
= ( a - 2b + 1 ) ( a - 1 )
a2 + 4ab + 3b2 - 2b - 1
= a2 + 4ab + 4b2 - b2 - 2b - 1
= ( a + 2b )2 - ( b - 1 ) 2
= ( a + 2b - b + 1 ) ( a + 2b + b -1 )
= ( a + b -1 ) ( a + 3b - 1)
Với a<2b<0, rút gọn \(\dfrac{1}{a-2b}\)√b2(a2-4ab+4b2)
\(\dfrac{1}{a-2b}.\sqrt{b^2\left(a^2-4ab+4b^2\right)}=\dfrac{1}{a-2b}.b.\left|a-2b\right|=\dfrac{1}{a-2b}.b.\left(2b-a\right)=-b\)
\(\dfrac{1}{a-2b}\cdot\sqrt{b^2\cdot\left(a^2-4ab+b^2\right)}\)
\(=\dfrac{1\cdot\left(a-2b\right)}{a-2b}\cdot b\)
=b
Rút gọn
A = 15(a + 2b)2 - 3(a + 2b)(a + 2b + 19) + 6(2a + 4b)(1 - a - 2b)
Đặt a + 2b = x
Ta có:
\(A=15x^2-3x\left(x+19\right)+12\left(1-x\right)\)
\(=15x^2-3x^2-57x+12x-12x^2\)
\(=-45x\)
\(=-45\left(a+2b\right)\)
Đặt a + 2b = x
Ta có:
\(A=15x^2-3x=\left(x+19\right)+12\left(1-x\right)\)
\(=15x^2-3x^2-57x+12x-12x^2\)
\(=-45x\)
\(=-45\left(a+2b\right)\)
2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc
2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
=2ab(a+2b)-ac(a+2b)+c2(a+2b)-2bc(a+2b)
=(a+2b)(2ab-ac+c2-2bc)
=(a+2b)\(\left[a\left(2b-c\right)-c\left(2b-c\right)\right]\)
=(a+2b)(2b-c)(a-c)
Rút gọn
a)\(2\sqrt{a}+3a\sqrt{4ab^2}-2b\sqrt{16a^5}-2\sqrt{25a}\)(a>0;b>0)
b)\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\left(a\ge0;b\ge0;a\ne b\right)\)
c)\(\frac{a\sqrt{a}-b\sqrt{b}}{a-b}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0;a\ne0\right)\)
Tìm điều kiện để các biểu thức sau có nghĩa và rút gọn chúng:
a. M=\(\sqrt{\frac{a^4b^3}{a^2b-ab}}\)
b.N= \(\frac{a}{b-1}.\frac{\sqrt{\left(b-1\right)^4}}{a^2}\)
Rút gọn:
\(A=15\left(a+2b\right)^2-3\left(a+2b\right)\left(a+2b+19\right)+6\left(2a+4b\right)\left(1-a-2b\right)\)
Rút gọn biểu thức: P = xy , biết ( 3 a 3 − 3 b 3 ) x − 2 b = 2 a với a ≠ b và ( 4 a + 4 b ) y = 9 ( a − b ) 2 với
Biến đổi được: x = 2 ( a + b ) 3 ( a 3 − b 3 ) ; y = 9 ( a − b ) 2 4 ( a + b )
⇒ P = x . y = 2 ( a + b ) 3 ( a 3 − b 3 ) . 9 ( a − b ) 2 4 ( a + b ) = 3 ( a − b ) 2 ( a 2 + ab + b 2 )