\(P=\dfrac{x^4+5x^3-20x^2-27x+30}{x^2+4x-21}\left(1\right)\)

Điều kiện xác định khi và chỉ khi

\(x^2+4x-21\ne0\)

\(\Leftrightarrow x^2+7x-3x-21\ne0\)

\(\Leftrightarrow x\left(x+7\right)-3\left(x+7\right)\ne0\)

\(\Leftrightarrow\left(x-3\right)\left(x+7\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-7\end{matrix}\right.\)

Theo đề bài : \(\)

\(x=\sqrt[]{31-12\sqrt[]{3}}=\sqrt[]{27-12\sqrt[]{3}+4}=\sqrt[]{\left(3\sqrt[]{3}-2\right)^2}=\left|3\sqrt[]{3}-2\right|=3\sqrt[]{3}-2\)

\(\left(1\right)\Leftrightarrow P=\dfrac{x^4-3x^3+8x^3-24x^2+4x^2-12x-15x+45-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3\left(x-3\right)+8x^2\left(x-3\right)+4x\left(x-3\right)-15\left(x-3\right)-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{\left(x-3\right)\left(x^3+8x^2+4x-15\right)-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+8x^2+4x-15}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+7x^2+x^2+7x-3x-15}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^2\left(x+7\right)+x\left(x+7\right)-3\left(x+7\right)+6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{\left(x^2+x-3\right)\left(x+7\right)+6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=x^2+x-3+\dfrac{6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

Thay \(x=3\sqrt[]{3}-2\) vào \(P\) ta được

\(\Leftrightarrow P=\left(3\sqrt[]{3}-2\right)^2+3\sqrt[]{3}-2-3+\dfrac{6}{3\sqrt[]{3}-2+7}-\dfrac{15}{\left(3\sqrt[]{3}-2-3\right)\left(3\sqrt[]{3}-2+7\right)}\)

\(\Leftrightarrow P=31-12\sqrt[]{3}+3\sqrt[]{3}-5+\dfrac{6}{3\sqrt[]{3}+5}-\dfrac{15}{\left(3\sqrt[]{3}-5\right)\left(3\sqrt[]{3}+5\right)}\)

\(\Leftrightarrow P=26-9\sqrt[]{3}+\dfrac{6\left(3\sqrt[]{3}-5\right)}{\left(3\sqrt[]{3}+5\right)\left(3\sqrt[]{3}-5\right)}-\dfrac{15}{\left(3\sqrt[]{3}\right)^2-5^2}\)

\(\Leftrightarrow P=26-9\sqrt[]{3}+\dfrac{6\left(3\sqrt[]{3}-5\right)}{2}-\dfrac{15}{2}\)

\(\Leftrightarrow P=\dfrac{37}{2}-9\sqrt[]{3}+3\left(3\sqrt[]{3}-5\right)\)

\(\Leftrightarrow P=\dfrac{37}{2}-9\sqrt[]{3}+9\sqrt[]{3}-15\)

\(\Leftrightarrow P=\dfrac{37}{2}-15=\dfrac{7}{2}\)

P = 7/2

TÍNH : \(\left(\sqrt{2}-1\right)^2-\frac{3}{2}\sqrt{\left(-2\right)^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.\sqrt{2}\)

\(=\left(\sqrt{2}-1\right)^2-\frac{3}{2}.2+\frac{4\sqrt{2}}{5}+\sqrt{\frac{36}{25}}.\sqrt{2}\)

\(=3-2\sqrt{2}-3+\frac{4\sqrt{2}}{5}+\frac{6\sqrt{2}}{5}=\frac{10\sqrt{2}}{5}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0\)

CHỨNG MINH : 

Ta có : \(\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right]+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)với mọi \(x\ge0\)

Vậy ta có điều phải chứng minh.

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

Lời giải:

Ta có:

\(\sqrt{x+4}+2\sqrt{x+1}=\sqrt{x+20}\)

\(\Leftrightarrow (\sqrt{x+4}+2\sqrt{x+1})^2=x+20\)

\(\Leftrightarrow 5x+8+4\sqrt{(x+1)(x+4)}=x+20\)

\(\Leftrightarrow \sqrt{(x+1)(x+4)}=3-x\)

Từ đây ta suy ra \(x\leq 3\)

Bình phương hai vế tiếp tục:

\(\Rightarrow (x+1)(x+4)=(3-x)^2\)

\(\Leftrightarrow x^2+5x+4=x^2-6x+9\Leftrightarrow 11x=5\)

\(\Leftrightarrow x=\frac{5}{11}\) (thử lại thấy thỏa mãn đkđb)

a: Đặt \(x^2-4=a\)

Pt sẽ là \(a=3\sqrt{xa}\)

\(\Rightarrow a^2=9xa\)

\(\Leftrightarrow a\left(a-9x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)

hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)

d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)

Pt sẽ là 2a+b=ab+2

=>(b-2)(1-a)=0

=>b=2 và 1-a

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

\(x=1\)