Tìm x
a) x ∈ BC( 3,5 ) và x < 50
b) x ⋮ 4; x ⋮ 6 và x < 40
c) x ⋮ 12; x⋮ 15 và x < 130
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a) \(\Leftrightarrow\left[{}\begin{matrix}x-3,5=7,5\\x-3,5=-7,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-4\end{matrix}\right.\)
b) \(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{2}\\x+\dfrac{4}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=-\dfrac{13}{10}\end{matrix}\right.\)
c) \(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
d) \(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\4,5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=4,5\end{matrix}\right.\)(vô lý)
Vậy \(S=\varnothing\)
\(x\in\left\{0;4;8;12;16;20;24;28;32;36;40;44;48\right\}\)
\(y\in\left\{1;2;4;13;26;52\right\}\)
a: x chia hết cho 4;5;10
nên \(x\in BC\left(4;5;10\right)\)
mà 10<=x<50
nên x=40
b: x=33
a, Ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
\(\Rightarrow x=11;y=17;z=23\)
b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)
\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
\(\Rightarrow x=6;y=9;z=15\)
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
xyz = 810
=> 2k.3k.5k = 810
=> k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
b: 30 chia hết cho x
45 chia hết cho x
Do đó: \(x\inƯC\left(30;45\right)=Ư\left(15\right)\)
mà x>10
nen x=15
c: \(\Leftrightarrow x+2\in\left\{1;-1;3;-3;9;-9\right\}\)
hay \(x\in\left\{-1;-3;1;-5;7;-11\right\}\)
d: =>x+3+14 chia hết cho x+3
=>\(x+3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)
hay \(x\in\left\{-2;-4;-1;-5;4;-10;11;-17\right\}\)
1,92x3,5<X<29,05:3,5
6,72<X<8,3
X=7 hoặc 8
Chúc bn học tốt
a) \(x\in BC\left(3,5\right)\Rightarrow x\in\left\{0;15;30;45;60;...\right\}\)
mà \(x< 50\Rightarrow x\in\left\{0;15;30;45\right\}\)
b) \(\hept{\begin{cases}x⋮4\\x⋮6\end{cases}\Rightarrow}x\in BC\left(4;6\right)=\left\{0;12;24;36;48;...\right\}\)
mà \(x< 40\Rightarrow x\in\left\{0;12;24;36\right\}\)
c) \(\hept{\begin{cases}x⋮12\\x⋮15\end{cases}\Rightarrow}x\in BC\left(12;15\right)=\left\{0;60;120;180;...\right\}\)
mà \(x< 130\Rightarrow x\in\left\{0;60;120\right\}\)
TL:
a) x∈BC(3,5)⇒x∈{0;15;30;45;60;...}
mà x<50⇒x∈{0;15;30;45}
b) {
⇒x∈BC(4;6)={0;12;24;36;48;...}
mà x<40⇒x∈{0;12;24;36}
c) {
⇒x∈BC(12;15)={0;60;120;180;...}
mà x<130⇒x∈{0;60;120}
^HT^