(x^2)^5= x^17/x^16
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5: \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
6: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
7: \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\)
8: \(x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\)
\(5,x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\\ 6,x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\\ 7,x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\\ 8,x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
Tính ngoặc tròn
\(4\cdot8-16\cdot2\)
\(=32-32\)
\(=0\)
Vậy tích trên bằng 0 ( vì có 1 thừa số = 0 )
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1
3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
9/17.x+268/17.x-345/17.x=16
(9/17+268/17-345/17).x=16
-4.x=16
x=16:-4
x=-4
\(\dfrac{9}{17}x+15\dfrac{13}{17}x-20\dfrac{5}{17}x=16\)
\(\left(\dfrac{9}{17}+15\dfrac{13}{17}-20\dfrac{5}{17}\right)x=16\)
\(\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)x=16\)
\(x=16:\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)\)
\(x=-4\)
Trả lời:
\(\left(x^2\right)^5=\frac{x^{17}}{x^{16}}\)
\(\Rightarrow x^{10}=x\)
\(\Rightarrow x^{10}-x=0\)
\(\Rightarrow x\left(x^9-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x = 0; x = 1