\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

Bài 4: 

c) Ta có: \(\dfrac{x^3}{8}+\dfrac{x^2y}{2}+\dfrac{xy^2}{6}+\dfrac{y^3}{27}\)

\(=\left(\dfrac{x}{2}\right)^3+3\cdot\left(\dfrac{x}{2}\right)^2\cdot\dfrac{y}{3}+3\cdot\dfrac{x}{2}\cdot\left(\dfrac{y}{3}\right)^2+\left(\dfrac{y}{3}\right)^3\)

\(=\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^3\)

\(=\left(\dfrac{-1}{2}\cdot8+\dfrac{1}{3}\cdot6\right)^3=\left(-4+2\right)^3=-8\)

Bài 6:

a) Ta có: \(P=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-12x\)

=0

b) Ta có: \(Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

=-8

:))))

c: Gọi bốn số nguyên liên tiếp là x;x+1;x+2;x+3

Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

\(=\left(x^2+3x+1\right)^2\)

\(d,M=\left(x^2-4xy+4y^2\right)-2\left(x-2y\right)+1+9\\ M=\left(x-2y\right)^2-2\left(x-2y\right)+1+9\\ M=\left(x-2y+1\right)^2+9\ge9\\ M_{min}=9\Leftrightarrow x=2y-1\)

\(a,a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\left(luôn.đúng.do.a+b+c=0\right)\)

a) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}\ne5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

Khi \(x=16\Rightarrow A=\dfrac{\sqrt[]{16}+2}{\sqrt[]{16}-5}=\dfrac{4+2}{4-5}=-6\)

b) \(B=\dfrac{3}{\sqrt[]{x}+5}+\dfrac{20-2\sqrt[]{x}}{x-25}\)

B có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{3\left(\sqrt[]{x}-5\right)+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{3\sqrt[]{x}-15+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{\sqrt[]{x}+5}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{1}{\sqrt[]{x}-5}\left(dpcm\right)\)

c) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\in Z\left(x\in Z\right)\)

\(\Leftrightarrow\sqrt[]{x}+2⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}+2-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}+2-\sqrt[]{x}+5⋮\sqrt[]{x}-5\)

\(\Leftrightarrow7⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}-5\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)

\(\Leftrightarrow x\in\left\{16;36;144\right\}\)

d) \(A>B\left(2\sqrt[]{x}+5\right)\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}>\dfrac{1}{\sqrt[]{x}-5}\left(2\sqrt[]{x}+5\right)\)

\(\Leftrightarrow\sqrt[]{x}+2>2\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}< -3\)

mà \(\sqrt[]{x}\ge0\)

\(\Leftrightarrow x\in\varnothing\)

C?

Câu “ Cuối tuần, nếu con được cô khen, mẹ sẽ thưởng cho con đi chơi công viên.” là: 

A. Câu kể             B. Câu hỏi            C. Câu khiến                D. Câu cảm

a

giúp mình câu c câu d với

\(P_1\) là góc nào?