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\(5^x+5^{x+2}=650;5^x.26=650;5^x=25;x=2\)
\(2^x+2^{x+3}=144;2^x.9=144;2^x=16;x=4\)
\(3^{x-1}+5.3^{x-1}=162;3^{x-1}.6=162;3^{x-1}=27;x=4\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\rightarrow x-5=0\&x-5=1\) hoặc x - 5 = - 1
\(x-5=1;x=6;x-5=0;x=5;x-5=-1;x=4\)
\(\left(2^2:4\right).2^n=4;2^n=2^2;n=2\)
B1. 2x + 3 + 22 = 72
=> 2x + 3 + 4 = 72
=> 2x + 3 = 72 - 4
=> 2x + 3 = 68
=> ko có gtri x
B2 : Ta có : A = 1 + 2 + 22 + 23 + 24 + 25 + 26 + ... + 22001 + 22002
= (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (22000 + 22001 + 22002)
= 3 + 22.(1 + 2 + 22) + 25.(1 + 2 + 22 ) + ... + 22000 . (1 + 2 + 22)
= 3 + 22.7 + 25.7 + ... + 22000 . 7
= 3 + (22 + 25 + .... + 22000) . 7
=> Số dư của 7 là 3
\(2^{10}.2^{x+4}=64^5\)
\(\Leftrightarrow2^{x+14}=2^{30}\)
\(\Leftrightarrow x+14=30\)
\(\Leftrightarrow x=16\)
\(5^x+5^{x+3}=630\)
\(\Rightarrow5^x.1+5^x.125=630\)
\(\Rightarrow5^x.126=630\)
\(\Rightarrow5^x=5\)
\(\Rightarrow x=1\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+..............+\left(x+100\right)=7450\)
\(\Rightarrow\left(x+x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=7450\)
\(101x+5050=7450\)
Đến đây tự tính
Bài 1 :
\(2^x.8=512\)
\(2^x=512:8\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
\(b,\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(c,x^{20}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(d,\left(x-3\right)^{10}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
Bài 1:
a,(2x-15):13+51=64
=> (2x-15):13=64-51
=> (2x-15):13=13
=>(2x-15)=1
=> 2x =16
=> x = 8
Vậy: x= 8
bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
+) \(A=3\left(x-4\right)^4-4\ge-4\)
Min A = -4 \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
+) \(B=5+2\left(x-2019\right)^{2020}\ge5\)
Min B = 5 \(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
+) \(C=5+2018\left(2020-x\right)^2\)
Min C = 5 \(\Leftrightarrow2020-x=0\Leftrightarrow x=2020\)
+) \(D=\left(x-1\right)^{2020}+\left(y+x\right)-1\ge-1\)
Min D = -1 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
+) \(E=2\left(x-1\right)^2+3\left(2x-y\right)^4-2\ge-2\)
Min E = -2 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
a, PT <=> \(1+8+27-16x=20\Leftrightarrow16x=16\Leftrightarrow x=1\)
b, \(x\left(x+1\right)-\left(x-2\right)=11\Leftrightarrow x^2+x-x+2=11\Leftrightarrow x^2-9=0\Leftrightarrow x=\pm3\)
Bài làm :
\(b,1^2+2^3+3^3-4^2.x=20\)
\(1+8+27-16.x=20\)
\(36-16.x=20\)
\(16x=36-20\)
\(16x=16\)
\(x=1\)
\(d,x.\left(x+1\right)-\left(x-2\right)=11\)
\(x^2+x-x+2=11\)
\(x^2+2=11\)
\(x^2=11-2\)
\(x^2=9\)
\(x^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Học tốt