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\(y=\frac{1}{x^2+\sqrt{x}}\frac{\left(\int^4_2^5^2_5_8\vec{\log_1\Rightarrow\beta}\right)}{100462}\)
(x-3):\(\frac{-7}{8}\)=\(\frac{-5}{2}\)-x
\(\frac{x-3}{\frac{-7}{8}}\)=\(\frac{-5}{2}\)-x
x-3=(\(\frac{-5}{2}\)-x).\(\frac{-7}{8}\)
x-3=\(\frac{35}{16}+\frac{7}{8}x\)
x-3-\(\frac{7}{8}x\)=\(\frac{35}{16}\)
\(\frac{x}{8}-3=\frac{35}{16}\)
\(\frac{x}{8}=\frac{83}{16}\)=\(\frac{2x}{16}\)
x=\(\frac{83}{2}\)
a) \(\left|\frac{4}{7}-x\right|+\frac{2}{5}=0\)
=> \(\left|\frac{4}{7}-x\right|=-\frac{2}{5}\), vô lí vì \(\left|\frac{4}{7}-x\right|\ge0\)
Vậy không tồn tại giá trị của x thỏa mãn đề bài
b) \(6-\left|\frac{1}{4}x+\frac{2}{5}\right|=0\)
=> \(\left|\frac{1}{4}x+\frac{2}{5}\right|=6-0=6\)
=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x+\frac{2}{5}=6\\\frac{1}{4}x+\frac{2}{5}=-6\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x=\frac{28}{5}\\\frac{1}{4}x=-\frac{32}{5}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)
c) \(\left|x-\frac{1}{3}\right|+\left|2-\frac{4}{5}\right|=0\)
=> \(\left|x-\frac{1}{3}\right|+\left|\frac{6}{5}\right|=0\)
=> \(\left|x-\frac{1}{3}\right|+\frac{6}{5}=0\)
=> \(\left|x-\frac{1}{3}\right|=-\frac{6}{5}\), vô lí vì \(\left|x-\frac{1}{3}\right|\ge0\)
Vậy không tồn tại giá trị của x thỏa mãn đề bài
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
3.(x-1/2) -5(x+3/5)=-x+1/5
3x - 3/2 -5x +3 = -x+1/5
3x-5x+x= 3/2-3+1/5
x.(3-5+1)=15/10 + (-30/10)+2/10
x.(-1)= -13/10
x = -13/10 : (-1)
x=13/10
vậy x=13/10
a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)
Ta có: \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\)
Dấu = xảy ra khi :
\(\frac{x}{5}+\frac{23}{2}=0\Leftrightarrow\frac{x}{5}=-\frac{23}{2}\Leftrightarrow x=-\frac{115}{2}\)
\(y-\frac{14}{3}=0\Leftrightarrow y=\frac{14}{3}\)
Vậy ..............
Ta có:
a) \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall y\)
=> \(\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{5}+\frac{23}{2}=0\\y-\frac{14}{3}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
Vậy Min của A = 2019 tại \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
câu b tượng tự
\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)
d) (2x+3)2016=(2x+3)2018 khi 2x+3=0 hoặc 1
Nếu 2x+3=0
=2x=-3 ( loại )
Nếu 2x+3=1
=>2x=-2
=>x=-1 ( thỏa )