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\(a,\frac{-3}{2}-2x+\frac{3}{4}=-1\)
\(\frac{-3}{2}-2x=-1-\frac{3}{4}\)
\(\frac{-3}{2}-2x=\frac{-7}{4}\)
\(2x=\frac{-7}{4}+\frac{-3}{2}\)
\(2x=\frac{-13}{4}\)
\(x=\frac{-13}{4}:2\)
\(x=\frac{-13}{4}.\frac{1}{2}\)
\(x=\frac{-13}{8}\)
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{9}{15}-\frac{10}{15}=\frac{-1}{15}\)
\(x=\frac{-1}{15}.\frac{1}{3}\)
\(x=\frac{-1}{45}\)
Vậy x = \(\frac{-1}{45}\)
c) \(\left|2x-1\right|+1=4\)
\(\left|2x-1\right|=4-1=3\)
2x-1 = 3 ; -3
TH1: 2.x - 1 = 3
2.x = 3 + 1 = 4
x = 4 : 2 = 2
TH2: 2.x - 1 = -3
2.x = -3 + 1 = -2
x = -2 : 2 = -1
Vậy x \(\in\){ 2 ; -1 }
Ngại làm ấn máy ==
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
1) \(5\left|x-3\right|=\frac{5}{7}\)
\(\left|x-3\right|=\frac{5}{7}:5\)
\(\left|x-3\right|=\frac{5}{7}.\frac{1}{5}=\frac{1}{7}\)
+) \(x-3=\frac{1}{7}\)
\(x=\frac{1}{7}+3=\frac{22}{7}\)
+) \(x-3=-\frac{1}{7}\)
\(x=-\frac{1}{7}+3=\frac{20}{7}\)
2 ) 3) tương tự
Bài 3:
a: \(S=1+5^2+5^4+\cdots+5^{200}\)
=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)
=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)
=>24S=\(5^{202}-1\)
=>\(S=\frac{5^{202}-1}{24}\)
b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)
\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)
mà \(4^{15}>3^{11}\)
nên \(4^{30}>3\cdot24^{10}\)
=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)
Bài 2:
a: |2x-3|>5
=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)
c: |3x-1|<=7
=>-7<=3x-1<=7
=>-6<=3x<=8
=>\(-2\le x\le\frac83\)
d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)
TH1: \(x<-\frac32\)
=>2x+3<0; 3x-5<0
(1) sẽ trở thành: -2x-3-3x+5=7
=>-5x+2=7
=>-5x=5
=>x=-1(loại)
TH2: -3/2<=x<5/3
=>2x+3>=0; 3x-5<0
(1) sẽ trở thành: 2x+3-3x+5=7
=>-x+8=7
=>-x=-1
=>x=-1(nhận)
TH3: x>=5/3
=>2x+3>0; 3x-5>=0
(1) sẽ trở thành: 2x+3+3x-5=7
=>5x-2=7
=>5x=9
=>x=9/5(nhận)
a, \((\frac{3}{7}-\frac{2}{3})\) .x =\(\frac{10}{21}\)
\(\frac{-5}{21}\).x=\(\frac{10}{21}\)
x= -2
Mk chỉ làm 1 phần các phằn còn lại tương tự



\(3x-\dfrac{3}{5}=\dfrac{-7}{10}\\ 3x=\dfrac{-7}{10}+\dfrac{3}{5}\\ 3x=\dfrac{-7}{10}+\dfrac{6}{10}\\ 3x=\dfrac{-1}{10}\\ x=\dfrac{-1}{30}\)
___________________
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\\ \dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\\ \dfrac{1}{3}:x=\dfrac{9}{15}-\dfrac{10}{15}\\ \dfrac{1}{3}:x=\dfrac{-1}{15}\\ x=\dfrac{1}{3}:\dfrac{-1}{15}\\ x=-5\)
\(3x-\dfrac{3}{5}=-\dfrac{7}{10}\)
\(3x\) \(=-\dfrac{7}{10}+\dfrac{3}{5}\)
\(3x\) \(=-\dfrac{1}{10}\)
\(x\) \(=-\dfrac{1}{10}:3\)
\(x\) \(=-\dfrac{1}{30}\)
Vậy \(x=-\dfrac{1}{30}\)
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\)
\(\dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{15}\)
\(x=\dfrac{1}{3}:-\dfrac{1}{15}\)
\(x=-5\)
Vậy \(x=-5\)
\(\)-5