\(\sqrt{4a^2+12a+9}+\sqrt{4a^2-12a+9}\) với \(-\dfrac{3}{2}\le a\le\dfrac{3}{2}\)

\(\sqrt{\left(2a+3\right)^3}+\sqrt{\left(2a-3\right)^3}\)

\(\left|2a+3\right|+\left|2a-3\right|\)

\(2a+3-2a+3\)

\(6\)

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 )

Viết rõ đề bài ra đc không ạ

đấy là phân số

Bài làm:

a) đkxđ: \(2a\ne\pm b\)

Ta có: \(M=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right)\div\left(\frac{1+4a^2+4b^2}{4a^2-b^2}\right)\)

\(M=\left[\frac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right].\left(\frac{\left(2a-b\right)\left(2a+b\right)}{4a^2+4b^2+1}\right)\)

\(M=\frac{4a+2b-6b-8a+4b}{4a^2+4b^2+1}\)

\(M=\frac{-4a}{4a^2+4b^2+1}\)

b) +Nếu: \(a=\frac{1}{3}\)và \(b=2\)

Khi đó GT của M là: \(M=\frac{-4.\frac{1}{3}}{4.\frac{1}{3^2}+4.2^2+1}=-\frac{12}{157}\)

Viết rõ đề ra nhá

\(P=\dfrac{9\sqrt{a}-\sqrt{25a}+\sqrt{4a^3}}{a^2+2a}=\dfrac{9\sqrt{a}-5\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{4\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{2\sqrt{a}\left(2+a\right)}{a\left(2+a\right)}=\dfrac{2\sqrt{a}}{a}=\dfrac{2.\sqrt{a}}{\sqrt{a}.\sqrt{a}}=\dfrac{2}{\sqrt{a}}\)

a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

a) \(\sqrt[]{1-4a+4a^2}\)

\(=\sqrt[]{\left(1-2a\right)^2}\)

\(=\left|1-2a\right|\)

\(=\left[{}\begin{matrix}1-2a\left(a\le\dfrac{1}{2}\right)\\2a-1\left(a>\dfrac{1}{2}\right)\end{matrix}\right.\)

b) \(x-2y-\sqrt[]{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt[]{\left(x-2y\right)^2}\)

\(=x-2y-\left|x-2y\right|\)

\(=\left[{}\begin{matrix}x-2y-x+2y\left(x\ge2y\right)\\x-2y+x-2y\left(x< 2y\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}0\left(x\ge2y\right)\\2x-4y\left(x< 2y\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}0\left(x\ge2y\right)\\2\left(x-2y\right)\left(x< 2y\right)\end{matrix}\right.\)

\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)

Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)

\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)

\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)

Với a > 0 và a ≠ 4 , ta có

T =   a −   2 a + 2   −   a +   2 a − 2   .   a   −   4 a   =    a −   2 2 −   a +   2 2   a −   2 . a + 2 .     a   − 4 a   =    a −   4 a   +   4 −   a −   4 a   −   4   a   − 4 .     a   − 4 a   =     − 8 a a   =   − 8