39+39*x=51012 tìm x
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Câu 1
a) \(48=2^4.3\)
\(60=2^2.3.5\)
\(72=2^3.3^2\)
\(ƯCLN\left(48;60;72\right)=2^2.3=12\)
\(ƯC\left(48;60;72\right)=Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
b) \(42=2.3.7\)
\(55=5.11\)
\(91=7.13\)
\(ƯCLN\left(42;55;91\right)=1\)
\(ƯC\left(42;55;91\right)=\left\{1\right\}\)
c) \(48=2^4.3\)
\(72=2^3.3^2\)
\(ƯCLN\left(48;72\right)=2^3.3=24\)
\(ƯC\left(48;72\right)=Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)
Câu 2:
120 ⋮ \(x\); 168 ⋮ \(x\); 216 ⋮ \(x\);
\(x\) \(\in\) ƯC(120; 168; 216)
120 = 23.3.5; 168 = 23.3.7; 216 = 23.33
ƯClN(120; 168; 216) = 23.3 = 24
\(x\) \(\in\) Ư(24) = {1; 2; 3; 4; 6; 8; 12; 24}
Vì \(x\) > 20 nên \(x\) = 24
Lời giải:
$y=\frac{x-3}{x+4}\Rightarrow y'=\frac{7}{(x+4)^2}; y''=\frac{-14}{(x+4)^3}$
\(A=2\left[\frac{7}{(x+4)^2}\right]^2+(1-\frac{x-3}{x+4}).\frac{-14}{(x+4)^3}\)
\(=\frac{98}{(x+4)^4}-\frac{98}{(x+4)^4}=0\)
\(=>x+39-39=x-39-39\)
\(x=x-39-39\)
\(x-78=x\)
\(x-78-x=x-x\)
\(0=-78\)
a) 35 × x = 805
x = 805 : 35
x = 23
b) x : 37 + 1238 = 1644
x : 37 = 1644 - 1238
x : 37 = 406
x = 406 × 37
x = 15022
Đề là \(f\left(x\right)=\dfrac{1}{2}sin2x-cosx-x+2015\) đúng không nhỉ?
\(f'\left(x\right)=cos2x+sinx-1\)
\(f'\left(x\right)=0\Leftrightarrow cos2x+sinx-1=0\)
\(\Leftrightarrow1-2sin^2x+sinx-1=0\)
\(\Leftrightarrow sinx\left(1-2sinx\right)=0\Rightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
a) \(5+3^{x+1}=86\)
\(=>3^{x+1}=86-5\)
\(=>3^{x+1}=81=3^4\)
\(=>x+1=4\) ( cùng cơ số )
\(=>x=4-1\)
\(=>x=3\)
b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(=>15:\left(x+2\right)=\left(27+3\right):10\)
\(=>15:\left(x+2\right)=30:10=3\)
\(=>x+2=15:3\)
\(=>x+2=5\)
\(=>x=5-2\)
\(=>x=3\)
c) \(\left(9x+2\right).4=80\)
\(=>9x+2=80:4\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:9\)
\(=>x=2\)
d) \(\left(245-x\right)+7^2=14\)
\(=>\left(245-x\right)+14=14\)
\(=>245-x=14-14\)
\(=>245-x=0\)
\(=>x=245-0\)
\(=>x=245\)
a: Để (d)//(d') nên \(\left\{{}\begin{matrix}4m-3=m+6\\m^2< >9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=3\\m\notin\left\{3;-3\right\}\end{matrix}\right.\Leftrightarrow m\in\varnothing\)
b: Để (d) trùng với (d') thì \(\left\{{}\begin{matrix}4m-3=m+6\\m^2=9\end{matrix}\right.\Leftrightarrow m=3\)
c: Để hai đường thẳng cắt nhau thì 4m-3<>m+6
hay m<>3
39*x=51012-39
39*x=50973
x=50973:39
x=1307
nhớ nha
39 + 39 * x = 51012
=> 39 * ( x + 1 ) = 51012
=> x + 1 = 1308
=> x = 1307