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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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\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Chúc bạn học tốt !!!
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2-\left(\frac{3}{5}\right)^2=0\)
Áp dụng bất đẳng thức A2 - B2 = ( A - B )( A + B )
Ta có :
\(\left(2x+\frac{3}{5}-\frac{3}{5}\right)\cdot\left(2x+\frac{3}{5}+\frac{3}{5}\right)=0\)
\(2x\left(2x+\frac{6}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x+\frac{6}{5}=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Vậy .......
(2x + 3/5) 2 -9/25 = 0
=) (2x + 3/5)2 = 9/25 = (+ - 3/5)2
=) 2x + 3/5 = 3/5 hoặc 2x + 3/5 = -3/5
=) 2x = 0 hoặc 2x = -6/5
=) x = 0 hoặc x = -3/5
bạn ghi lại đề nha
(2x+3/5)^2=9/25
=> 2x+3/5=3/5 hoặc 2x+3/5= -3/5
2x=0 2x= -6/5
=> x=0 x= -6/5:2
x=-3/5
\(\left[2x+\left(\frac{3}{5}\right)^2\right]-\frac{9}{25}=0\)
=> \(2x+\frac{9}{25}-\frac{9}{25}=0\)
=> \(2x+0=0\)
=> \(x=0\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=\frac{-3}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-\frac{12}{5}\end{cases}}}\)
Vậy \(x=\left\{0;\frac{-12}{5}\right\}\)
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
a) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
(2x+3/5)2-9/25=0
(2x+3/5)2 =0+9/25
(2x+3/5)2 =9/25
(2x+3/5)2 = (3/5)2
2x+3/5 =3/5
2x =3/5-3/5
2x =0
x =0:2
x =0