Tìm x, biết
60% x+0.4x+x:3=2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{3}{5}x+\frac{2}{5}x+\frac{x}{3}=2\)
\(x+\frac{x}{3}=2\)
\(\frac{4}{3}x=2\)
\(x=2:\frac{4}{3}=\frac{3}{2}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-9-3x-9}{2\left(x-9\right)}=\dfrac{-2x-18}{2\left(x-9\right)}=\dfrac{-x-9}{x-9}\)
\(\begin{array}{l} a)3x - 1 = x + 3\\ \Leftrightarrow 3x - x = 3 + 1\\ \Leftrightarrow 2x = 4\\ \Leftrightarrow x = 2\\ b)15 - 7x = 9 - 3x\\ \Leftrightarrow - 7x + 3x = 9 - 15\\ \Leftrightarrow - 4x = - 6\\ \Leftrightarrow x = \dfrac{3}{2}\\ c)x - 3 = 18\\ \Leftrightarrow x = 18 + 3\\ \Leftrightarrow x = 21 \end{array}\)
\(\begin{array}{l} d)2x + 1 = 15 - 5x\\ \Leftrightarrow 2x + 5x = 15 - 1\\ \Leftrightarrow 7x = 14\\ \Leftrightarrow x = 2\\ e)3x - 2 = 2x + 5\\ \Leftrightarrow 3x - 2x = 5 + 2\\ \Leftrightarrow x = 7\\ f) - 4x + 8 = 0\\ \Leftrightarrow - 4x = - 8\\ \Leftrightarrow x = 2 \end{array}\)
a, (-0,4x-2)0(1.5x+1)-(-4x-0.8)=3.6
<=>-0,4x-2-1,5x-1+4x+0.8=3,6
<=>2,1x+2,2=3,6
<=>2,1x=1,4
<=>x=2/3
0,4x2 - 4x -1200 =0
\(\Delta=b^2-4ac\) = (-4)2 - 4. (1200) . 0,4 = 1936
=> pt có 2 nghiệm phân biệt.
x1= \(\dfrac{-\left(-4\right)-\sqrt{1936}}{0,4}\)= 100
x2=\(\dfrac{-\left(-4\right)+\sqrt{1936}}{0,4}\)= 120
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 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: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
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)
\(60\%x+0,4.x+\frac{x}{3}=2\)
\(\frac{3}{5}.x+\frac{2}{5}.x+\frac{1}{3}.x=2\)
\(x.\left(\frac{3}{5}+\frac{2}{5}+\frac{1}{3}\right)=2\)
\(x.\frac{4}{3}=2\)
\(x=2:\frac{4}{3}\)
\(x=\frac{3}{2}=1,5\)
60% x + 0,4x + x : 3 = 2
3/5 . x + 2/5 . x + x : 3 = 2
(3/5 + 2/5) x + x : 3 = 2
x + x : 3 = 2
x : (1 + 3) = 2
x : 4 = 2
x = 8