\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)

\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)

\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)

a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)

b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)

\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)

\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)

c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)

\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)

\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)

\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)

a.\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ac-2\left(a^2+2ab+b^2\right)=2a^2+2b^2+2c^2+4ab-2a^2-2ab-2b^2=2c^2+2ab\)

b. \(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\left(a^2+b^2-c^2+a^2-b^2+c^2\right)=\left(2b^2-2c^2\right).2a^2=4a^2\left(b^2-c^2\right)=4a^2b^2-4a^2c^2\)

\(a,=x^3-16x-x^2-1-x^2+1=x^3-2x^2-16x\\ b,=y^4-81-y^4+4=-77\\ d,=a^2+b^2+c^2+2ab-2bc-2ac+a^2-2ac+c^2-2ab-2ac\\ =2a^2+b^2+2c^2-2bc-6ac\)

\(A=\dfrac{n!+2}{\dfrac{n!}{\left(n-k\right)!}\cdot n!-k}+\dfrac{3003+10010+6435}{19448}\)

\(=\dfrac{n!+2}{n\left(n-1\right)\cdot...\cdot\left(n-k+1\right)\cdot n!-k}+1=\dfrac{n!+2+\dfrac{n!^2}{\left(n-k\right)!}-k}{\dfrac{n!^2}{\left(n-k\right)!}-k}\)

\(B=\dfrac{n!-\left(n-1\right)!}{\left(n-2\right)!}=\dfrac{\left(n-1\right)!\left(n-1\right)}{\left(n-2\right)!}=\left(n-1\right)^2=n^2-2n+1\)

\(log_{a^4}b^4.log_ba^5=\dfrac{1}{4}.4.log_ab.5.log_ba=5.log_ab.log_ba=5\)

\(log_{a^3}b^2.log_ba^4=\dfrac{1}{3}.2.log_ab.4.log_ba=\dfrac{8}{3}.log_ab.log_ba=\dfrac{8}{3}\)

\(log_{a^{15}}b^7.log_{b^{49}}a^{30}=\dfrac{1}{15}.7.log_ab.\dfrac{1}{49}.30.log_ba=\dfrac{2}{7}log_ab.log_ba=\dfrac{2}{7}\)

\(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}=\dfrac{1}{2021}.2020.log_ab.\dfrac{1}{4040}.6063.log_ba=\dfrac{3}{2}\)

a: \(A=\dfrac{3}{5}\cdot xy\cdot\dfrac{4}{25}x^2y^4z^2=\dfrac{12}{125}x^3y^5z^2\)

b: Hệ số là 12/125

Bậc là 10

A = (x –y)²+ 4xy
= x²-2xy+y²+4xy
=  x²⁺2xy+y²
=(x+y)²
B = (a + b)²+ (a –b)²
=(a+b+a-b)(a+b-a+b)
=2a.2b
=4ab

 

\(A=\left(x-y\right)^2+4xy=\left(x+y\right)^2\)

\(B=\left(a+b\right)^2+\left(a-b\right)^2=2a^2+2b^2\)

a: \(A=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}-1}{a-b}\)

\(=\dfrac{\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}+1}{a-b}=\dfrac{1}{a-b}\)

b: Khi a-b=1 thì A=1/1=1

\(a,\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\left(dkxd:a\ne9,a\ge0\right)\)

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

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

\(=-\sqrt{a}+3+3-\sqrt{a}\)

\(=6-2\sqrt{a}\)

\(b,\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\left(dkxd:a\ne b,a\ge0,b\ge0\right)\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)

\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=0\)

a) \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\)

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

\(=\dfrac{3-\sqrt{a}}{1}-\dfrac{\sqrt{a}-3}{1}\)

\(=3-\sqrt{a}-\sqrt{a}+3\)

\(=-2\sqrt{a}+6\)

b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{\sqrt{a}-\sqrt{b}}{1}-\dfrac{\sqrt{a}-\sqrt{b}}{1}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=0\)

`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

=(x+1)^2-2(2x-1)(x+1)+(2x-1)^2

=(x+1-2x+1)^2

=(-x+2)^2=x^2-4x+4