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3 tháng 8 2022

Vì (x-1/2)^2 lớn hơn hoặc bằng 0 với mọi x thuộc R

Mà theo bài:  (x-1/2)^2 = -4

Suy ra: phương trình vô nghiệm

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

24 tháng 4 2022

chưa biết

5 tháng 9 2023

1) \(S=2.2.2..2\left(2023.số.2\right)\)

\(\Rightarrow S=2^{2023}=\left(2^{20}\right)^{101}.2^3=\overline{....6}.8=\overline{.....8}\)

2) \(S=3.13.23...2023\)

Từ \(3;13;23;...2023\) có \(\left[\left(2023-3\right):10+1\right]=203\left(số.hạng\right)\)

\(\) \(\Rightarrow S\) có số tận cùng là \(1.3^3=27\left(3^{203}=\left(3^{20}\right)^{10}.3^3\right)\)

\(\Rightarrow S=\overline{.....7}\)

3) \(S=4.4.4...4\left(2023.số.4\right)\)

\(\Rightarrow S=4^{2023}=\overline{.....4}\)

4) \(S=7.17.27.....2017\)

Từ \(7;17;27;...2017\) có \(\left[\left(2017-7\right):10+1\right]=202\left(số.hạng\right)\)

\(\Rightarrow S\) có tận cùng là \(1.7^2=49\left(7^{202}=7^{4.50}.7^2\right)\)

\(\Rightarrow S=\overline{.....9}\)

a: =>(2x+1/2)^2=1/4

=>2x+1/2=1/2 hoặc 2x+1/2=-1/2

=>x=-1/2 hoặc x=0

b: =>(x-1/5)^2=49

=>x-1/5=7 hoặc x-1/5=-7

=>x=-6,8 hoặc x=7,2

c: =>1,2x=12

=>x=10

d: =>3/4x+1/2x+1/2=-11/4

=>5/4x=-11/4-2/4=-13/4

=>x=-13/5

e: =>-0,25x+1,25x=0,2

=>x=0,2

`@` ` \text {Ans}`

`\downarrow`

`a,`

`1/4+3/4*x=3/2-x`

`=> 1/4 + 3/4x - 3/2 + x = 0`

`=> (1/4 - 3/2) + (3/4x + x) = 0`

`=> -5/4 + 7/4x = 0`

`=> 7/4x = 5/4`

`=> x = 5/4 \div 7/4`

`=> x = 5/7`

Vậy, `x=5/7`

`b,`

`3/5*x-1/4=1/10*x-1/2`

`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`

`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`

`=> 1/2x + 1/4 = 0`

`=> 1/2x = -1/4`

`=> x = -1/4 \div 1/2`

`=> x = -1/2`

Vậy, `x=-1/2`

`c,`

`3x-3/5=x-1/4`

`=> 3x - 3/5 - x + 1/4 = 0`

`=> (3x - x) - (3/5 - 1/4) = 0`

`=> 2x - 7/20 = 0`

`=> 2x = 0,35`

`=> x = 0,35 \div 2`

`=> x = 7/40`

Vậy, `x=7/40`

`d,`

`3/2*x-2/5=1/3*x-1/4`

`=>  3/2x - 2/5 - 1/3x + 1/4 = 0`

`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`

`=> 7/6x - 3/20 = 0`

`=> 7/6x = 3/20`

`=> x = 3/20 \div 7/6`

`=> x = 9/70`

Vậy, `x=9/70`

`@` `\text {Kaizuu lv uuu}`

17 tháng 8 2023

a)

`x+1/4xx1/5=4/5`

`x+1/20=4/5`

`x=4/5-1/20`

`x=16/20-1/20`

`x=15/20=3/4`

b)

`3 2/3:x=2 2/3-1 1/2`

`11/3:x=8/3-3/2`

`11/3:x=16/6-9/6`

`11/3:x=7/6`

`x=11/3:7/6`

`x=11/3xx6/7`

`x=22/7`

17 tháng 8 2023

\(a,x+\dfrac{1}{4}.\dfrac{1}{5}=\dfrac{4}{5}\)

\(\Rightarrow\)\(x+\dfrac{1}{20}=\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{4}{5}-\dfrac{1}{20}=\dfrac{16}{20}-\dfrac{1}{20}=\dfrac{3}{4}\)

\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)

\(\Rightarrow\dfrac{11}{3}:x=\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)

22 tháng 8 2023

\(a,x+\dfrac{1}{4}\times\dfrac{1}{5}=\dfrac{4}{5}\)

\(x+\dfrac{1}{20}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}-\dfrac{1}{20}\)

\(x=\dfrac{16}{20}-\dfrac{1}{20}\)

\(x=\dfrac{15}{20}=\dfrac{3}{4}\)

\(b,3\dfrac{2}{3}:x=2\dfrac{2}{3}-1\dfrac{1}{2}\)

\(\dfrac{11}{3}:x=\dfrac{8}{3}-\dfrac{3}{2}\)

\(\dfrac{11}{3}:x=\dfrac{7}{6}\)

\(x=\dfrac{11}{3}:\dfrac{7}{6}=\dfrac{22}{7}\)

a: =>x+1/20=4/5

=>x=4/5-1/20=16/20-1/20=15/20=3/4

b: =>11/3:x=8/3-3/2=16/6-9/6=7/6

=>x=11/3:7/6=11/3*6/7=66/21=22/7

5 tháng 3 2022

\(x-\dfrac{1}{2}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}+\dfrac{1}{2}\)

\(x=\dfrac{5}{4}\)

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}-\dfrac{7}{8}\)

\(x=\dfrac{-1}{8}\)

\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)

\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)

\(x=\dfrac{-1}{2}\)

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