\(E=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}\)

\(=\frac{1}{2}.\frac{3}{4}\)

\(=\frac{3}{8}\)

Ta có: A = 25 - |3x - 6| - |3x + 8|

A = 25 - (|6 - 3x| + |3x + 8|) < = 25 - |6 - 3x + 3x + 8| = 25 - |14| = 25 - 14 = 11

Dấu "=" xảy ra <=> (3x - 6)(3x + 8) = 0

=> -8/3 \(\le\)x \(\le\)2

Vậy Max của A = 11 tại \(-\frac{8}{3}\le x\le2\)

Ta có: B = |2x - 5| - |2x - 11| + 3 > = |2x - 5 - 2x + 11| + 3 = |6| + 3  = 6 + 3 = 9

Dấu "=" xảy ra <=> (2x - 5)(2x - 11) = 0

=> \(\frac{5}{2}\le x\le\frac{11}{2}\)

Vậy Min của B = 9 tại \(\frac{5}{2}\le x\le\frac{11}{2}\)

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

b) \(x^2-3x+2=0\Leftrightarrow x^2-3x+\frac{9}{4}-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}=\sqrt{\frac{1}{4}}=\frac{1}{2}\\x-\frac{3}{2}=-\sqrt{\frac{1}{4}}=-\frac{1}{2}\end{cases}}\)

Giải tiếp nha

Ta co:

\(\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\)

Dau "=" xay ra khi:

\(1\le x\le5\)

\(\left|y+1\right|\ge0\Rightarrow\left|y+1\right|+3\ge3\Rightarrow\frac{12}{\left|y+1\right|+3}\le4\)

Dau "=" xay ra khi:

\(\left|y+1\right|=0\Leftrightarrow y=-1\)

Ma \(\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}\)

\(\Rightarrow\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}=4\)

Vay \(1\le x\le5;y=-1\)

\(4x^4+4x^2=0\)

\(\Leftrightarrow4x^2\left(x^2+1\right)=0\)

\(\Leftrightarrow x=0\)

\(3x^4+4x^2=0\)

\(x^4\left(3+4\right)=0\)

=>\(x^4=0\)

=>\(x=0\)

Ta có: \(4x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{4}\Leftrightarrow\frac{x}{45}=\frac{y}{36}\)

\(14x=9z\Leftrightarrow\frac{x}{9}=\frac{z}{14}\Leftrightarrow\frac{x}{45}=\frac{z}{70}\)

\(\Leftrightarrow\frac{x}{45}=\frac{y}{36}=\frac{z}{70}=\frac{2x}{90}=\frac{3y}{108}\)

Áp dụng t/c dãy tỉ số = nhau, ta có:

\(\frac{2x}{90}=\frac{3y}{108}=\frac{2x-3y}{90-108}=\frac{-10}{-18}=\frac{5}{9}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{45}=\frac{5}{9}\\\frac{y}{36}=\frac{5}{9}\\\frac{z}{70}=\frac{5}{9}\end{cases}\Rightarrow}\hept{\begin{cases}x=25\\y=20\\z=\frac{350}{9}\end{cases}}\)

Ta có :

\(4x=5y\Rightarrow4x=5y=\frac{y}{4}=\frac{x}{5}\)

\(14x=9z\Rightarrow14x=9z=\frac{z}{14}=\frac{x}{9}\)

VẬY NÊN ta có :     \(\frac{y}{4}=\frac{x}{5},\frac{x}{9}=\frac{z}{14}\Rightarrow\frac{y}{36}=\frac{x}{45},\frac{x}{45}=\frac{z}{70}\)

\(\Rightarrow\frac{y}{36}=\frac{x}{45}=\frac{z}{70}\)

ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU TACÓ :

\(\frac{X-Y}{36-45}\)=\(\frac{2X-3Y}{72-135}=\frac{-10}{-63}\)

MÌNH CHỈ LÀM ĐẾN ĐÓ THÔI DÀI LẮM

a) Thay x = 1 vào M(x), ta được:

\(M\left(x\right)=m.1^2+2m.1-6=m+2m-6=3m-6=0\)

\(\Leftrightarrow3m=6\Leftrightarrow m=2\)

Vậy m = 2 thì M(x) có nghiệm bằng 1

\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)

\(\Leftrightarrow\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}-\frac{x-4}{2016}=0\)

\(\Leftrightarrow\frac{x-1}{2019}-1+\frac{x-2}{2018}-1-\frac{x-3}{2017}+1-\frac{x-4}{2016}+1=0\)

\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)

\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)

\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\frac{x-1}{2019}+\frac{x-2}{2018}-2=\frac{x-3}{2017}+\frac{x-4}{2016}-2\)

\(\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

\(\frac{x-1-2019}{2019}+\frac{x-2-2018}{2018}=\frac{x-3-2017}{2017}+\frac{x-4-2016}{2016}\)

\(\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)

\(\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)

\(\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

\(\Rightarrow x-2020=0\)

Vậy \(x=2020\)