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Ta có : 3a2 + 2b2 = 7ab ( a > b > 0 )
⇔ 3a2 - 6ab - ab + 2b2 = 0
⇔ 3a( a - 2b) - b( a - 2b) = 0
⇔ ( a - 2b)( 3a - b) = 0
⇔ a = 2b ( TM ĐK ) hoặc 3a = b ( KTM ĐK)
Khi đó : \(A=\dfrac{a^3-b^3}{\left(a+b\right)ab}=\dfrac{\left(2b-b\right)\left(4b^2+2b^2+b^2\right)}{3b.2b^2}=\dfrac{7b^3}{6b^3}=\dfrac{7}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$7-3a=(\sqrt{7}-\sqrt{3a})(\sqrt{7}+\sqrt{3a})$
b.
$14x^2-11=(\sqrt{14}x-\sqrt{11})(\sqrt{14}x+\sqrt{11})$
c.
$3x-6\sqrt{x}-6=3(x-2\sqrt{x}-2)$
$=3[(\sqrt{x}-1)^2-3]$
$=3(\sqrt{x}-1-\sqrt{3})(\sqrt{x}-1+\sqrt{3})$
d.
$x\sqrt{x}-3\sqrt{x}-2=x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2$
$=x(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}-2)+(\sqrt{x}-2)$
$=(\sqrt{x}-2)(x+2\sqrt{x}+1)$
$=(\sqrt{x}-2)(\sqrt{x}+1)^2$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(7ab\cdot\sqrt{\dfrac{36a^4}{49b^2}}\\ =>7ab\cdot\sqrt{\left(\dfrac{6a^2}{7b}\right)^2}\\ =>7ab\cdot\dfrac{6a^2}{7b}\\ =>\dfrac{7ab\cdot6a^2}{7b}\\ =>6a^3\)
\(A=7ab.\sqrt{\dfrac{36a^4}{49b^2}}\)
\(=7ab.\dfrac{6a^2}{\left|7b\right|}\)
\(=7ab.\dfrac{6a^2}{7b}\left(vib>0\right)\)
\(=6a^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3a-2\sqrt{ab}-b=3a-3\sqrt{ab}+\sqrt{ab}-b\)
\(=3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)=\left(3\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
b) \(5a+3\sqrt{ab}-8b=5a-5\sqrt{ab}+8\sqrt{ab}-8b\)
\(=5\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+8\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(5\sqrt{a}+8\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
a) (\(\sqrt{a}-\sqrt{b}\))(3\(\sqrt{a}+b\))
b) \(\left(\sqrt{a}-\sqrt{b}\right)\left(5\sqrt{a}+8\sqrt{b}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)
c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)
d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)+1\)
\(=\left(a+b-1\right)^2\)
Câu 2:
Xét BToán \(x+y+z=0\Leftrightarrow x^3+y^3+z^3=3xyz\)
Mà \(\left(x-y\right)+\left(y-z\right)+\left(z-x\right)=0\)
\(\Rightarrow\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(a^2+3a=2\)
\(b^2+3b=2\)
=> \(\left(a-b\right)\left(a+b\right)+3\left(a-b\right)=0\)
=> \(\left(a-b\right)\left(a+b+3\right)=0\)
=> a = b ( loại ) hoặc a + b = - 3 ( Thỏa mãn )
Ta có : \(a^2+3a=2\Rightarrow a^3=2a-3a^2\)
\(b^2+3b=2\Rightarrow b2b-3b^2\)
=> \(a^3+b^3=2a+2b-3\left(2-3a\right)-3\left(2-3b\right)\)
\(=11\left(a+b\right)-12=11\left(-3\right)-12=-45\)
Đa thức này không phân tích được thành nhân tử bạn nhé. Bạn xem lại đề.