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a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
Bài 1:
b) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\frac{1}{2}\\x=0+\frac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{4}\right\}.\)
c) \(\left(2x-5\right)^4=81\)
\(\Rightarrow2x-5=\pm3\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3+5=8\\2x=\left(-3\right)+5=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8:2\\x=2:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{4;1\right\}.\)
d) \(3^{x+1}+3^{x+3}=810\)
\(\Rightarrow3^x.3^1+3^x.3^3=810\)
\(\Rightarrow3^x.\left(3^1+3^3\right)=810\)
\(\Rightarrow3^x.30=810\)
\(\Rightarrow3^x=810:30\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a) (2x+1)4=34
--> 2x+1=3 hay 2x+1=-3
-->2x=2 hay 2x=-4
--> x=1 hay x=-2
b) (x-1/2)3= (-2)3
--> x-1/2=-2
--> x=1/2-2
x=-3/4
c) (x-5)x+1=(x-5)13
--> x+1=13
x=13-1
x=12