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 \(x=2020\)

\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\Rightarrow\)\(\frac{x+32}{11}-3+\frac{x+23}{12}-2=\frac{x+38}{13}-3+\frac{x+27}{14}-2\)

\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\Rightarrow\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}-\frac{x-1}{14}=0\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Rightarrow x-1=0\)(Vì \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\))

\(\Rightarrow x=1\)

Vậy:x=1

\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)) 

\(\Leftrightarrow x=-1\)

Vậy x=-1

mỗi phân số + 1 thì sẽ có tử chung là x + 1

chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\))  =0

mà tổng các phân số kia khác 0 nên x+1 bằng 0 

=> x=-1

Viet lai de bai

Cho \(\frac{a}{b}=\frac{c}{d}\)

CMR:\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)

Bai lam:

Dat \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Ta co:

\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)

\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)

\(\frac{x}{10}=\frac{y}{5}\Rightarrow\frac{x}{20}=\frac{y}{10}\)

\(\frac{y}{2}=\frac{z}{3}\Rightarrow\frac{y}{10}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{20}=\frac{y}{10}=\frac{z}{15}\)

\(\Rightarrow\frac{2x}{40}=\frac{3y}{30}=\frac{z}{15}\)

áp dụng tc của dãy tỉ số = nhau

=> x/20=y/10; y/10=z/15

=> x/20=y/10=z/15

từ...áp dụng....

đc : (2x-3y+4z)/(40-30+60)=280/70=4

  => x=..

=> y=...

=> z=...

bạn tự làm nha