Tìm x: x+2*x+3*x=606
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a)83-6.(x-7)=35
6(x-7)=83-35=48
x-7=48:6=8
x=8+7=15
b)246+7.(x+15)=442
7(x+15)=442-246=196
x+15=196:7=28
x=28-15=13
a.
a,550 : 5 x 8
=110 x 8
=880
b,946 - 606 : 3
=946-202
=744
c,(285 + 60) : 5 x 8
=345:5x8
=69 x8
=552
d,735 : (3 + 2) - 50
=147 -50
=97
e,105 + 82 : 2
=105+41
=146
g,420 x 2 - 120
=840-120
=720
h,6 x (25 + 125)
=6 x150
=900
i,150 + 25 x 3
=150+75
=225
a. 550 : 5 x 8 = 880. b. 946 - 606 : 3 = 643. c. (285 + 60) : 5 x 8 = 552. d. 735 : (3 + 2) - 50 = 97.
e. 105 + 82 : 2 = 146. g. 420 x 2 - 120 = 720. h. 6 x (25 + 125) = 900. i. 150 + 25 x 3 = 225.
a:
Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)
\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)
\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)
b: x^2-4x+3=0
=>x=1(nhận) hoặc x=3(loại)
Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)
c: P>0
=>x-2>0
=>x>2
d: P nguyên
=>4x^2 chia hết cho x-2
=>4x^2-16+16 chia hết cho x-2
=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}
=>x thuộc {1;4;6;-2;10;-6;18;-14}
Bài 3:
a) Đặt f(x)=0
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) Đặt f(x)=0
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Bài 3:
c) Đặt f(x)=0
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
d) Đặt f(x)=0
\(\Leftrightarrow x^4+2=0\)
\(\Leftrightarrow x^4=-2\)(Vô lý)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
X + 2 x X + 3 x X = 606
X x ( 1 + 2 + 3 ) = 606
X x 6 = 606
X = 606 : 6
X = 101
\(x+2\) x \(x+3\) x \(x=606\)
\(x+2\) x \(x+3\) x \(x+1=606\)
\(x\) x \(\left(2+3+1\right)=606\)
\(x\) x \(6=606\)
\(x=606:6\)
\(x=101\)