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27 tháng 6 2016

oho

12 tháng 7 2023

Mày nhìn cái chóa j

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

26 tháng 2 2022

hic, mk chx học

28 tháng 2 2020

Ta có vế trái của pt luôn \(\ge0\)

Do đó : \(11x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\...\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{cases}}\)

Khi đó pt trở thành :

\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)

\(\Leftrightarrow10x+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=11x\)

\(\Leftrightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Leftrightarrow x=1-\frac{1}{11}=\frac{10}{11}\) ( thỏa mãn )

Vậy : pt đã cho có nghiệm \(S=\left\{\frac{10}{11}\right\}\)

28 tháng 2 2020

Dễ thấy \(VT>0\forall x\)

\(\Rightarrow11x>0\Rightarrow x>0\)

Phương trình trở thành \(10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow x=1-\frac{1}{11}=\frac{10}{11}\)

Vậy \(x=\frac{10}{11}\)

2 tháng 7 2020

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

2 tháng 7 2020

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)

Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)

\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)

\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)

\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)

\(\Leftrightarrow18x-9=0\)

\(\Leftrightarrow9\left(2x-1\right)=0\)

mà 9≠0

nên 2x-1=0

⇔2x=1

hay \(x=\frac{1}{2}\)(tm)

Vậy: \(x=\frac{1}{2}\)

b)ĐKXĐ: x≠0

Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)

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

\(\Leftrightarrow x^3+x-x^4-1=0\)

\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)

\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)

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

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

Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)

Từ (1) và (2) suy ra x-1=0

hay x=1(tm)

Vậy: x=1

c) ĐKXĐ: x≠0

Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)

\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)

Ta có: 1≠0

x≠0

Do đó: \(\frac{1}{x}\ne0\)

\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)

Từ (3) và (4) suy ra x=0(ktm)

Vậy: x∈∅

d) ĐKXĐ: x≠0

Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)

\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)

\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)

\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)

\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)

mà 4≠0

và x≠0

nên \(1+\frac{1}{x}=0\)

\(\Leftrightarrow\frac{1}{x}=-1\)

hay x=-1(tm)

Vậy: x=-1

24 tháng 4 2019

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

24 tháng 4 2019

ĐKXĐ: x khác -4;-5;-6;-7

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow3.18=x^2+11x+28\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right).\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}\left(tm\right)}\)

Vậy...