(x+\(\dfrac{3}{2}\))2=\(\dfrac{9}{49}\)
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a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a: Áp dụng tính chất của DTSBN, ta được:
x/5=y/2=(x-y)/(5-2)=9/3=3
=>x=15; y=6
b: =>(x-3)/12=3/(x-3)
=>(x-3)^2=36
=>(x-9)(x+3)=0
=>x=9 hoặc x=-3
c; x/2=y/3
=>x/10=y/15
y/5=z/4
=>y/15=z/12
=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17
=>x=490/17; y=735/17; z=588/17
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
a) Ta có: \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
hay \(x=\dfrac{3}{4}\)
b) Ta có: \(\left(x+\dfrac{4}{9}\right)^2=\dfrac{49}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=\dfrac{7}{12}\\x+\dfrac{4}{9}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{36}\\x=\dfrac{-37}{36}\end{matrix}\right.\)
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
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Nếu ko thì lấy máy tính mà tính cũng đc mà
\(\dfrac{3}{14}\cdot\left(-49\right)=-\dfrac{21}{2}\)
\(\dfrac{3}{4}\cdot\dfrac{-18}{25}=-\dfrac{27}{50}\)
\(1\dfrac{2}{3}\cdot3\dfrac{2}{9}=\dfrac{29}{9}\cdot\dfrac{5}{3}=\dfrac{145}{27}\)
\(40\%\cdot\dfrac{20}{9}=\dfrac{40}{100}\cdot\dfrac{20}{9}=\dfrac{40}{45}=\dfrac{8}{9}\)
a, \(-49.\dfrac{3}{14}=-\dfrac{21}{2}\)
b, \(\dfrac{-18}{25}.\dfrac{3}{4}=-\dfrac{27}{50}\)
c, \(3\dfrac{2}{9}.1\dfrac{2}{3}=\dfrac{29}{9}.\dfrac{5}{3}=\dfrac{145}{27}\)
d, \(\dfrac{20}{9}.40\%=\dfrac{20}{9}.\dfrac{40}{100}=\dfrac{8}{9}\)
\(\left(x+\dfrac{3}{2}\right)^2=\dfrac{9}{49}\)
\(\left(x+\dfrac{3}{2}\right)^2=\left(\pm\dfrac{3}{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{3}{7}\\x+\dfrac{3}{2}=\dfrac{-3}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-15}{14}\\x=\dfrac{-27}{14}\end{matrix}\right.\)
Vậy \(x=\dfrac{-15}{14}\) hoặc \(x=\dfrac{-27}{14}\) thỏa mãn .
tik mik nha !!!
\(\left(x+\dfrac{3}{2}\right)^2=\left(\dfrac{3}{7}\right)^2\Leftrightarrow x+\dfrac{3}{2}=\dfrac{3}{7}\Leftrightarrow x=\dfrac{-15}{14}\)