a: \(A=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)

\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\)

<=>\(\dfrac{4\left(7x-2\right)}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3\left(x-2\right)}{12}\)

<=>\(4\left(7x-2\right)-24x< 60-3\left(x-2\right)\)

<=>\(28x-8-24x< 60-3x+6\)

<=>\(28x+3x-24x< 60+8+6\)

<=>\(7x< 74\)

<=>x<\(\dfrac{74}{7}\)

Vậy...

Câu 1:

c. 

PT $(1)\Leftrightarrow x=1+2my$. Thay vô PT $(2)$:
$m(1+2my)+y=2$
$\Leftrightarrow y(2m^2+1)=2-m$

$\Leftrightarrow y=\frac{2-m}{2m^2+1}$

$x=1+2my=1+\frac{4m-2m^2}{2m^2+1}=\frac{4m+1}{2m^2+1}$
Vậy hpt có nghiệm duy nhất $(x,y)=(\frac{4m+1}{2m^2+1}, \frac{2-m}{2m^2+1})$
Để $x,y$ nguyên thì:
$4m+1\vdots 2m^2+1$ và $2-m\vdots 2m^2+1$

$\Rightarrow 4m+1+4(2-m)\vdots 2m^2+1$
$\Leftrightarrow 9\vdots 2m^2+1$

$\Rightarrow 2m^2+1\in\left\{1;3;9\right\}$

$\Rightarrow m\in\left\{0; 1; -1;2;-2\right\}$

Thử lại thì thấy $m=0; -1;2$ thỏa mãn.

\(\frac{a}{2}=\frac{b}{5}=\frac{c}{8}\)

\(=>\frac{a}{2}=\frac{b}{5}=\frac{2c}{16}\)

ÁP DỤNG T/C DÃY TỈ SỐ BẰNG NHAY, TA CÓ:

\(\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{2c}{16}=\frac{a-b+2c}{2-5+16}=\frac{6}{13}\)

\(\frac{a}{2}=\frac{6}{13}=>a=\frac{12}{13}\)

\(\frac{b}{5}=\frac{6}{13}=>b=\frac{30}{13}\)

\(\frac{c}{8}=\frac{6}{13}=>c=\frac{48}{13}\)

Vậy a=...

b=...

c=...

b: B=418-3253+253+582

=1000-3000

=-2000

1: Bảng giá trị:

Vẽ đồ thị:

2: Phương trình hoành độ giao điểm là: \(-\frac12x^2=-x-4\)

=>\(x^2=2x+8\)

=>\(x^2-2x-8=0\)

=>(x-4)(x+2)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-2\end{array}\right.\)

Khi x=4 thì y=-x-4=-4-4=-8

Khi x=-2 thì y=-x-4=-(-2)-4=2-4=-2

=>I(4;-8); K(-2;-2)

A thuộc tia Ox nên A(x;0)

(d): y=-x-4

=>y+x+4=0

=>x+y+4=0

KHoảng cách từ A đến (d) là:

\(d\left(A;\left(d\right)\right)=\frac{\left|x\cdot1+0\cdot1+4\right|}{\sqrt{1^2+1^2}}=\frac{\left|x+4\right|}{\sqrt2}\)

I(4;-8); K(-2;-2)

=>\(IK=\sqrt{\left(-2-4\right)^2+\left(-2+8\right)^2}=\sqrt{\left(-6\right)^2+6^2}=6\sqrt2\)

Diện tích tam giác AIK là \(15\operatorname{cm}^2\) nên ta có:

\(S_{AIK}=\frac12\cdot IK\cdot d\left(A;IK\right)\)

=>\(\frac12\cdot6\sqrt2\cdot\frac{\left|x+4\right|}{\sqrt2}=15\)

=>\(3\left|x+4\right|=15\)

=>|x+4|=5

=>\(\left[\begin{array}{l}x+4=5\\ x+4=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5-4=1\\ x=-5-4=-9\end{array}\right.\)

=>A(1;0) hoặc A(-9;0)

c: \(\Leftrightarrow x^2-5x-x^2-7< =0\)

=>-5x<=7

hay x>=-7/5

d: \(\Leftrightarrow x^2-x-2+3-x^2>=0\)

=>-x+1>=0

=>-x>=-1

hay x<=1

a) Ta có: \(\sqrt{2018}-\sqrt{2017}=\dfrac{1}{\sqrt{2018}+\sqrt{2017}}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

mà \(\dfrac{1}{\sqrt{2018}+\sqrt{2017}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

nên \(\sqrt{2018}-\sqrt{2017}< \sqrt{2016}-\sqrt{2015}\)

b) Ta có: \(\left(\sqrt{2018}+\sqrt{2016}\right)^2=4034+24\sqrt{14126}\)

\(\left(2\sqrt{2017}\right)^2=8068=4034+4034\)

mà \(24\sqrt{14126}< 4034\)

nên \(\sqrt{2018}+\sqrt{2016}< 2\sqrt{2017}\)