1.a.

\(\left(x+1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)\ge m\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-10\right)\ge m\)

Đặt \(x^2+3x-10=t\ge-\dfrac{49}{4}\)

\(\Rightarrow\left(t+2\right)t\ge m\Leftrightarrow t^2+2t\ge m\)

Xét \(f\left(t\right)=t^2+2t\) với \(t\ge-\dfrac{49}{4}\)

\(-\dfrac{b}{2a}=-1\) ; \(f\left(-1\right)=-1\) ; \(f\left(-\dfrac{49}{4}\right)=\dfrac{2009}{16}\)

\(\Rightarrow f\left(t\right)\ge-1\)

\(\Rightarrow\) BPT đúng với mọi x khi \(m\le-1\)

Có 30 giá trị nguyên của m

1b.

Với \(x=0\)  BPT luôn đúng

Với \(x\ne0\) BPT tương đương:

\(\dfrac{\left(x^2-2x+4\right)\left(x^2+3x+4\right)}{x^2}\ge m\)

\(\Leftrightarrow\left(x+\dfrac{4}{x}-2\right)\left(x+\dfrac{4}{x}+3\right)\ge m\)

Đặt \(x+\dfrac{4}{x}-2=t\) \(\Rightarrow\left[{}\begin{matrix}t\ge2\\t\le-6\end{matrix}\right.\)

\(\Rightarrow t\left(t+5\right)\ge m\Leftrightarrow t^2+5t\ge m\)

Xét hàm \(f\left(t\right)=t^2+5t\) trên \(D=(-\infty;-6]\cup[2;+\infty)\)

\(-\dfrac{b}{2a}=-\dfrac{5}{2}\notin D\) ; \(f\left(-6\right)=6\) ; \(f\left(2\right)=14\)

\(\Rightarrow f\left(t\right)\ge6\)

\(\Rightarrow m\le6\)

Vậy có 37 giá trị nguyên của m thỏa mãn

2.

Xét với \(x\ge1\)

\(m\left(x+1\right)+3\left(x-1\right)-2\sqrt{x^2-1}=0\)

\(\Leftrightarrow m+3\left(\dfrac{x-1}{x+1}\right)-2\sqrt{\dfrac{x-1}{x+1}}=0\)

Đặt \(\sqrt{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow m+3t^2-2t=0\)

\(\Leftrightarrow3t^2-2t=-m\)

Xét hàm \(f\left(t\right)=3t^2-2t\) trên \(D=[0;1)\)

\(-\dfrac{b}{2a}=\dfrac{1}{3}\in D\) ; \(f\left(0\right)=0\) ; \(f\left(\dfrac{1}{3}\right)=-\dfrac{1}{3}\) ; \(f\left(1\right)=1\)

\(\Rightarrow-\dfrac{1}{3}\le f\left(t\right)< 1\)

\(\Rightarrow\) Pt có nghiệm khi \(-\dfrac{1}{3}\le-m< 1\)

\(\Leftrightarrow-1< m\le\dfrac{1}{3}\)

\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)

1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)

\(B=\dfrac{1}{2018}\)

2)a)\(x^2-2x-15=0\)

\(\Leftrightarrow x^2-2x+1-16=0\)

\(\Leftrightarrow\left(x-1\right)^2-16=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

3)\(\dfrac{a}{b}=\dfrac{d}{c}\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)

Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)

\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)

4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)

\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)

\(g\left(x\right)=-x^{101}+f\left(x\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)

Tại x=0 thì f(x)-g(x)=0

Tại x=1 thì f(x)-g(x)=1

CHu làm cô liễu ko lo làm Mai báo cô

a.

Do \(0\le x\le1\Rightarrow\left(1+x\right)^2\ge\left(x+x\right)^2=4x^2\) (đpcm)

Dấu "=" xảy ra khi \(x=1\)

b.

Do \(x;y\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\) \(\Rightarrow x+y\ge x^2+y^2\)

\(\Rightarrow\left(1+x+y\right)^2\ge4\left(x+y\right)\ge4\left(x^2+y^2\right)\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)

a, \(A\cup B=(-4;5]\)

\(A\cap B=[-3;4)\)

\(A\backslash B=\left[4;5\right]\)

\(B\backslash A=\left(-4;-3\right)\)

b, \(A\cup B=\left(-3;7\right)\)

\(A\cap B=[1;2)\cup(3;5]\)

\(A\backslash B=\left[2;3\right]\)

\(B\backslash A=\left(-3;1\right)\cup\left(5;7\right)\)

c, \(A\cup B=\left[\dfrac{1}{2};3\right]\)

\(A\cap B=\left[1;\dfrac{3}{2}\right]\)

\(A\backslash B=[\dfrac{1}{2};1)\)

\(B\backslash A=(\dfrac{3}{2};3]\)

d, \(A\cup B=(-5;2]\cup(3;6]\)

\(A\cap B=\left\{0\right\}\cup[4;5)\)

\(A\backslash B=(0;2]\cup\left[-5;6\right]\)

\(B\backslash A=[-5;0)\cup\left(3;4\right)\)

a) -5 + |3x - 1| + 6 = |-4|

=> -5 + |3x - 1| + 6 = 4

=> 1 + |3x - 1| = 4

=> |3x - 1| = 4 - 1

=> |3x - 1| = 3

=> \(\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)

=> \(\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)

=>  \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy ...

d) |x + 1| + |x + 2| + |x + 3| = 4x

Ta có: |x + 1| \(\ge\)0  \(\forall\)x

          |x + 2| \(\ge\)0 \(\forall\)x

          |x + 3| \(\ge\)0 \(\forall\)x

=> |x + 1| + |x + 2| + |x + 3| \(\ge\)0 \(\forall\)x => 4x \(\ge\)0 \(\forall\) x=> x \(\ge\)0 \(\forall\)x

=> x + 1 + x + 2 + x + 3 = 4x

=> 3x + 6 = 4x

=> 6 = 4x - 3x

=> x = 6

Vậy...

b) (x - 1)² = (x - 1)⁴

=> (x - 1)² - (x - 1)⁴ = 0

=> (x - 1)² .[1 - (x - 1)² ] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\)

=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy x = {1; 2; 0}