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Câu hỏi của chíp chíp - Toán lớp 7 | Học trực tuyến Full dễ ok??
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\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
a: =>5/42-x=11/13-15/28+11/13=421/364
=>x=-1193/1092
b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)
=>2x=1/2
=>x=1/4
c: =>|2x-1/3|=-1/3(vô lý)
d: =>2x-1=-3
=>2x=-2
hay x=-1
e: =>2x=16
hay x=8
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55