Ôn thi vào 10

`B=(2+[5-\sqrt{5}]/[\sqrt{5}-1]).(2-[5+\sqrt{5}]/[\sqrt{5}+1])`

`B=(2+[\sqrt{5}(\sqrt{5}-1)]/[\sqrt{5}-1]).(2-[\sqrt{5}(\sqrt{5}+1)]/[\sqrt{5}+1])`

`B=(2+\sqrt{5})(2-\sqrt{5})`

`B=4-5=-1`

\(P=\left(\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)+3x+6\sqrt{x}}{\sqrt{x}\left(x-4\right)}\right):\dfrac{x\sqrt{x}-2\sqrt{x}+2x-4-x}{\sqrt{x}\left(x-2\right)}\)

\(=\dfrac{4x-2\sqrt{x}-8}{\sqrt{x}\left(x-4\right)}\cdot\dfrac{\sqrt{x}\left(x-2\right)}{x\sqrt{x}+x-2\sqrt{x}-4}\)

\(=\dfrac{4x-2\sqrt{x}-8}{x-4}\cdot\dfrac{x-2}{x\sqrt{x}+x-2\sqrt{x}-4}\)

b: Tọa độ điểm A là:

\(\left\{{}\begin{matrix}x=0\\y=3\cdot0+2=2\end{matrix}\right.\Leftrightarrow A\left(0;2\right)\)

Tọa độ điểm B là:

\(\left\{{}\begin{matrix}y=0\\3x+2=0\end{matrix}\right.\Leftrightarrow B\left(-\dfrac{2}{3};0\right)\)

\(OA=\sqrt{\left(0-0\right)^2+\left(2-0\right)^2}=2\)

\(OB=\sqrt{\left(-\dfrac{2}{3}-0\right)^2}=\dfrac{2}{3}\)

\(S_{OAB}=\dfrac{OA\cdot OB}{2}=\dfrac{4}{3}:2=\dfrac{4}{6}=\dfrac{2}{3}\)

*Xét \(n=11k\left(k\in N\right)\):

\(A=n^2+9n-2=121k^2+99k-2⋮̸11\)(loại).

*Xét \(n=11k+1\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+22k+1+99k+9-2=121k^2+22k+99k+8⋮̸11\)(loại)

*Xét \(n=11k+2\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+44k+4+99k+18-2=121k^2+44k+99k+20⋮̸11\)(loại)

*Xét \(n=11k+3\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+66k+9+99k+27-2=121k^2+66k+99k+34⋮̸11\)

(loại)

*Xét \(n=11k+4\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+88k+16+99k+36-2=121k^2+88k+99k+50⋮̸11\)

(loại)

*Xét \(n=11k+5\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+110k+25+99k+45-2=121k^2+110k+99k+68⋮̸11\)(loại)

*Xét \(n=11k+6\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+132k+36+99k+54-2=121k^2+132k+99k+88⋮11\)

(nhận)

*Xét \(n=11k+7\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+154k+49+99k+63-2=121k^2+154k+99k+110⋮11\)

(nhận) 

*Xét \(n=11k+8\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+176k+64+99k+72-2=121k^2+176k+99k+134⋮̸11\)

(loại)

*Xét \(n=11k+9\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+198k+81+99k+81-2=121k^2+198k+99k+160⋮̸11\)

(loại)

*Xét \(n=11k+10\left(k\in N\right)\)

\(A=n^2+9n-2=121k^2+220k+100+99k+90-2=121k^2+220k+99k+188⋮̸11\)

(loại)

Tổng kết lại, vậy với mọi n nguyên dương chia 11 dư 6 hoặc 7 thì biểu thức A chia hết cho 11.

\(=\left(3-\sqrt{5}+\dfrac{8}{\sqrt{5}-1}\right).\dfrac{1}{\sqrt{5}+1}=\dfrac{3-\sqrt{5}}{\sqrt{5}+1}+2=\dfrac{3-\sqrt{5}+2\sqrt{5}+2}{\sqrt{5}+1}=\dfrac{\sqrt{5}+5}{\sqrt{5}+1}=\sqrt{5}\)

\(x^{x^2-5x+6}=1\)

\(\Rightarrow x^{x^2-5x+6}=1\)

dựa vào công thức tính nghiệm , ta tính được nghiệm sau:

x₁ = -1

x₂ = 1 ; x₃ = 2 ; x₄ = 3

Thử lại nghiệm,ta thu được tập nghiệm của phương trình : \(x^{x^2-5x+6}=1\) là \(\left\{1;2;3;-1\right\}\)

\(A=\Sigma\dfrac{1}{a^5\left(b+2c\right)^2}=\Sigma\dfrac{\left(abc\right)^3}{a^5\left(b+2c\right)^2}=\Sigma\dfrac{b^3c^3}{a^2\left(b+2c\right)^2}=\Sigma\dfrac{b^3c^3}{\left(ab+2ac\right)^2}\)

\(có:\dfrac{b^3c^3}{\left(ab+2ac\right)^2}+\dfrac{ab+2ac}{27}+\dfrac{ab+2ac}{27}\ge3\sqrt[3]{\dfrac{b^3c^3}{27^2}}=\dfrac{bc}{3}\)

\(tương\) \(tự\Rightarrow A\ge\left(\dfrac{bc+ac+ab}{3}\right)-\left(\dfrac{2\left(ab+2ac\right)+2\left(bc+2ab\right)+2\left(ca+2bc\right)}{27}\right)\)

\(=\dfrac{9\left(ab+bc+ac\right)-2\left(ab+2ac+bc+2ab+ca+2bc\right)}{27}=\dfrac{3\left(ab+bc+ca\right)}{27}\ge\dfrac{3}{27}.3\sqrt[3]{\left(abc\right)^2}=\dfrac{1}{3}\)

\(dấu"="\Leftrightarrow a=b=c=1\)