x-2tất cả trên 8 =17/2
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\(A=\frac{\left(4^2\right)^{17}.64^{36}}{8^{35}.32^{34}}\)
\(A=\frac{\left(2^4\right)^{17}.\left(2^6\right)^{36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=\frac{2^{68}.2^{216}}{2^{105}.2^{170}}=\frac{2^{284}}{2^{275}}=2^9=512\)
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 ...
1) ở đây : https://hoc24.vn/hoi-dap/question/637404.html
2) pt \(\Leftrightarrow\left(9x^2+18x-16\right)\left(x^2+2x+1\right)=-16\)
\(\Leftrightarrow9x^4+36x^3+29x-14x=0\)
\(\Leftrightarrow x\left(9x^3+18x^2+18x^2+36x-7x-14\right)=0\)
\(\Leftrightarrow x\left(9x^2\left(x+2\right)+18x\left(x+2\right)-7\left(x+2\right)\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(9x^2+18x-7\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(9x^2-3x+21x-7\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(3x\left(3x-1\right)+7\left(3x-1\right)\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(3x+7\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=\dfrac{-7}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\) vậy ............................................................................................................
a) Liệt kê
x = {-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7}
Tính tổng là: -7+-6+-5+-4+.....+4+5+6+7
= (-7+7)+(-6+6)+(-5+5)+....+(-1+1)+0
= 0+0+0....+0
= 0
b) Liệt kê
x = {-5;-4;-3;-2;-1;0;1;2;3}
Tính tổng: -5+-4+-3+-2+-2+0+1+2+3
= (-3+3)+(-2+2)+(-1+1)+0+-5+-4
= 0+0+0+0+ -9
= -9
c) Liệt kê:
x = { -19;-18;-17;-16;....;18;19;20}
Tính tổng: -19+-18+-17+-16+....+15+16+17+18+19+20
= (-19+19)+(-18+18)+...+(-1+1)+0+20
= 0 + 0+...+0+20
= 20
*TÌM X:
a) 2x -35 = 15
2x = 15 + 35
2x = 50
x = 50 :2
x = 25
b) 3x + 17 = 2
3x = 17+2
3x = 19
x = 19 : 3
x = 6,33
c) /x-1/ = 0
\(\hept{\begin{cases}x-1=0\\x-1=-0\left(loai\right)\end{cases}}\)
Vậy x-1 = 0
x = 0 +1 = 1
Câu 1: Tìm x:
a) 2/5× x =(thiếu đề)
8/7: x =4/5
x=8/7:4/5
x=10/7
Câu 2: Tính bằng cách thuận tiện nhất
5/9× 8/17 + 4/9 × 8/17
=(5/9+4/9)x8/17
=1x8/17=8/17
ý bạn là vầy phải k:
\(\frac{x-2}{8}=\frac{17}{2}\)
\(\frac{x-2}{8}=\frac{17}{2}\)
\(\left(x-2\right).2=8.17\)
\(\left(x-2\right).2=126\)
\(x-2=126:2\)
\(x-2=63\)
\(x=63+2\)
\(x=65\)