2I5x-3I-2x=14
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a) 2|5x - 3| - 2x = 14
=> 2|5x - 3| = 14 + 2x
=> |5x - 3| = x + 7
=> \(\orbr{\begin{cases}5x-3=x+7\\5x-3=-x-7\end{cases}}\)
=> \(\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
=> \(\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}\)(tm)
\(\Rightarrow2^{x+1}=16=2^4\\ \Rightarrow x+1=4\\ \Rightarrow x=3\)
Bài 1:
a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)
b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)
c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)
d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)
e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)
Bài 2:
\(40-x=2^6\cdot2^2\)
\(\Rightarrow40-x=2^8\)
\(\Rightarrow40-x=256\)
\(\Rightarrow x=40-256\)
\(\Rightarrow x=-216\)
b) \(3^2\cdot3^x=81\)
\(\Rightarrow3^{2+x}=3^4\)
\(\Rightarrow2+x=4\)
\(\Rightarrow x=4-2=2\)
c) \(2^x=512\)
\(\Rightarrow2^x=2^9\)
\(\Rightarrow x=9\)
d) \(x^5=243\)
\(\Rightarrow x^5=3^5\)
\(\Rightarrow x=3\)
Bài 3:
a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)
b) \(8^3=\left(2^3\right)^3=2^9=512\)
c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)
d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)
e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)
\(=18-\left(4+3\right)\)
\(=18-7=11\)
2:
a: =>40-x=256
=>x=40-256=-216
b: =>x+2=4
=>x=2
c: =>2^x=2^9
=>x=9
d; =>x^5=3^5
=>x=3
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
= x2(5x-7)+2(5x-7)
= (5x-7)(x2+2)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+3x+9+3x)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
=(5x3+10x)-(7x2+14)
=5x(x2+2)-7(x2+2)
=(x2+2)(5x-7)
\(x\times\dfrac{3}{5}=\dfrac{2}{3}+\dfrac{1}{3}\)
\(x\times\dfrac{3}{5}=\dfrac{3}{3}\)
\(x\times\dfrac{3}{5}=1\)
\(x=1:\dfrac{3}{5}\)
\(x=1\times\dfrac{5}{3}\)
\(x=\dfrac{5}{3}\)
\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)
\(x-\dfrac{4}{9}=\dfrac{3}{7}\times\dfrac{14}{9}\)
\(x-\dfrac{4}{9}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{4}{9}\)
`x`\(=\dfrac{18}{27}+\dfrac{12}{27}\)
\(x=\dfrac{30}{27}=\dfrac{10}{9}\)
Câu 1: Tác phẩm Quốc âm thi tập là của ai ? (1đ)
a. Nguyễn Trãi.
b. Lê Thánh Tông.
c. Lý Tử Tấn.
d .Lương Thế Vinh.
Ta có: \(2|5x-3|-2x=14\)
\(\Leftrightarrow2|5x-3|=14+2x\)
\(\Leftrightarrow|5x-3|=7+x\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-2}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)
\(2\left|5x-3\right|-2x=14\)
\(\Rightarrow2\left|5x-3\right|=14+2x\)
\(\Rightarrow\left|5x-3\right|=\left(14+2x\right):2=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-2}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)
Chúc em học tốt nhé!