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\(2\cdot2^2\cdot2^3\cdot2^4\cdot\cdot\cdot2^x=32768\)
\(\Leftrightarrow2^{1+2+3+4+\cdot\cdot\cdot+x}=2^{15}\)
\(\Leftrightarrow1+2+3+4+..+x=15\)
\(\Leftrightarrow\)\(\frac{\left(1+x\right)x}{2}=15\)
\(\Leftrightarrow x\left(x+1\right)=30=5\left(5+1\right)\)
Vậy x=5
Bài 2:
Bậc của đơn thức là 2+5+3=10
Bài 3:
\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=5\)
+)TH1: \(x\ge\frac{1}{4}\) thì bt trở thành
\(2x-\frac{1}{2}=5\Leftrightarrow2x=\frac{11}{2}\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)
+)TH2: \(x< \frac{1}{4}\) thì pt trở thành
\(2x-\frac{1}{2}=-5\Leftrightarrow2x=-\frac{9}{2}\Leftrightarrow x=-\frac{9}{4}\left(tm\right)\)
Vậy x={-9/4;11/4}
2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)
\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3
3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)
4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)
a: Số số hạng là \(\dfrac{2018-2}{2}+1=1009\left(số\right)\)
Tổng là: \(\dfrac{2018+2}{2}\cdot1009=1009\cdot1010=1019090\)
b: \(10S=10^2+10^3+...+10^{101}\)
\(\Rightarrow9S=10^{101}-10\)
hay \(S=\dfrac{10^{101}-10}{9}\)
c: \(5S=1+\dfrac{1}{5}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow4S=1-\dfrac{1}{5^{100}}\)
hay \(S=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)
\(4^x-10.2^x+16=0\)
\(\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)
Đặt 2x = t
\(\Rightarrow t^2-10t+16=0\)
\(\Leftrightarrow t^2-2t-8t+16=0\)
\(\Leftrightarrow t\left(t-2\right)-8\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t-8\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}t=2\\t=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2^x=2\\2^x=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(4^x-10\times2^x+16=0\)
\(\Leftrightarrow2^{2x}-2\times5\times2^x+16=0\)
\(\Leftrightarrow\left[\left(2^x\right)^2-2\times2^x\times5+25\right]-9=0\)
\(\Leftrightarrow\left(2^x-5\right)^2-3^2=0\)
\(\Leftrightarrow\left(2^x-5-3\right)\left(2^x-5+3\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-2\right)=0\)
\(\Leftrightarrow2^x-8=0\) hoặc \(2^x-2=0\)
\(\cdot2^x-8=0\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(\cdot2^x-2=0\Leftrightarrow2^x=2\Leftrightarrow x=1\)
Vậy \(S=\left\{3;1\right\}\)
\(4^x-10\cdot2^x+16=0\)
\(=\left(2^x\right)^2-10\cdot2^x+16=0\)
Đặt \(t=2^x\). Ta có:
\(t^2-10t+16=0\)
\(\Rightarrow t^2-2\cdot t\cdot5+25-9=0\)
\(\Rightarrow\left(t-5\right)^2-3^2=0\)
\(\Rightarrow\left(t-8\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=8\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy S = {1,3}
a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)
ĐK: \(x\ne2\).
a) \(P=\frac{x+1}{x-2}=\frac{x-2+3}{x-2}=1+\frac{3}{x-2}\)nguyên mà \(x\)nguyên nên \(x-2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\)
suy ra \(x\in\left\{-1,1,3,5\right\}\).
Thử lại để \(P\)nguyên dương thì \(x\in\left\{-1,3,5\right\}\).
b) \(-x^2-x+2=0\)
\(\Leftrightarrow-x^2+x-2x+2=0\)
\(\Leftrightarrow\left(-x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\Rightarrow P=\frac{1}{4}\\x=1\Rightarrow P=-2\end{cases}}\)
2x22x23.....x2x=32768=215
=>21+2+3+...+x=215 =>1+2+3+...+x=15
=>\(\dfrac{x\left(x+1\right)}{2}=15=>x\left(x+1\right)=30=5.\left(5+1\right)\)
=>x=5
bạn ơi cho tớ hỏi sao lại có x(x+1) /2 ..... vì sao phải chia 2 bạn.
kq đúng ròi bạn
Đặt A=1+2+3+...+x
<=>A+A=1+2+3+...+x+1+2+3+...+x
2A=(x+1)+(x-1+2)+(x-2+3)+....(x+1) (có x lần x+1)
2A=x(x+1) => A=x(x+1)/2