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3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
\(3^{x+2}+3^x=90\Leftrightarrow3^x.3^2+3^x=90\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\Leftrightarrow x=2\)
Vậy ...
\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)
Vậy ...
1.
a) \(3^{x+2}+3^x=90\)
\(\Leftrightarrow3^x\left(3^2+1\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
vậy...
b) \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
Vậy...
tik mik nha !!!
a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow3\left(x-2\right)=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=2x+80\)
\(\Leftrightarrow4x-2x=80-92\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)
d, \(B=1+2+2^2+........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)
\(< =>3x-6=5x=>x=-3\)
\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)
\(4x+92=3x+120=>x=28\)
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
\(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow\left(x-2\right)3=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=-6\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy .....
b, \(B=1+2+2^2+..........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=3x+120\)
\(\Leftrightarrow4x-3x=120-92\)
\(\Leftrightarrow x=28\)
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
(1/3+x)2=4/9
(1/3+x)2=(2/3)2
TH1
1/3+x=2/3
x=2/3-1/3
x=1/3
TH2
1/3+x=-2/3
x=-2/3-1/3
x=-1
Vậy x bằng 1/3 và -1