tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

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3(3x-1/2)^3+1/9=0

2/3-1/3x(X-3/2)-1/2x(2X+1)=5

2/3-1/3+1/2-x-1/2=5

5/6-x-1/2=5

5/6-5-1/2=x

x=-25/6-1/2

x=-14/3

mình nhanh nhất li ke cho mình nha

\(x\div4\frac{1}{3}=-2,5\)

\(\Leftrightarrow x\div\frac{13}{3}=\frac{-5}{2}\)

\(\Leftrightarrow x=\frac{-5}{2}.\frac{13}{3}\)

\(\Leftrightarrow x=\frac{-65}{6}\)

\(x\div\frac{-3}{5}=\frac{-10}{21}\)

\(\Leftrightarrow x=\frac{-10}{21}.\frac{-3}{5}\)

\(\Leftrightarrow x=\frac{30}{105}\)

\(\Leftrightarrow x=\frac{2}{7}\)

\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}+\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{6}{10}\)

\(\Leftrightarrow x=\frac{6}{10}\div\frac{2}{3}\)

\(\Leftrightarrow x=\frac{18}{20}\)

\(\Leftrightarrow x=\frac{9}{10}\)

\(\frac{1}{2}x+\frac{1}{2}=\frac{5}{2}\)

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(x+1\right)=\frac{5}{2}\div\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)=5\)

\(\Leftrightarrow x=5-1\)

\(\Leftrightarrow x=4\)

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)