![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
PT $\Leftrightarrow 0,4+0,2x-0,5x=0,25-0,5x+0,25$
$\Leftrightarrow 0,4-0,3x=0,5-0,5x$
$\Leftrightarrow 0,2x=0,1\Rightarrow x=0,5$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
\(\Leftrightarrow\dfrac{2+x}{5}-\dfrac{x}{2}=\dfrac{1-2x}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2+x}{5}-\dfrac{x}{2}=\dfrac{1-2x+1}{4}\)
\(\Leftrightarrow\dfrac{2+x}{5}-\dfrac{x}{2}=\dfrac{2-2x}{4}\)
\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1-x}{2}+\dfrac{x}{2}\)
\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1-x+x}{2}\)
\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(2+x\right)=5\\ \Leftrightarrow2x+4-5=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\)
\(PT.\Rightarrow\) \(\dfrac{8+4x-10x-5+10x-5}{20}=0.\Rightarrow4x=2.\Leftrightarrow x=\dfrac{1}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
\(\Rightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Rightarrow8+4x-10x=5-10x+5\)
\(\Rightarrow8+4x=10\)
\(\Rightarrow4x=2\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy tập nghiệm của phương trình là A = { 1/2 }
\(\frac{8+4x-10x-5+10x}{20}=0,25\)
\(\frac{4x+3}{20}=0,25\)
\(4x+3=5\)
\(x=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
![](https://rs.olm.vn/images/avt/0.png?1311)
=>0,2x+0,4-0,5x=0,25-0,5x+0,25
=>0,2x+0,4=0,5
=>0,2x=0,1
=>x=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}=\dfrac{x}{6}=\dfrac{6x}{6}\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow2x-6x-3=x-6x\)
\(\Leftrightarrow2x-6x-x+6x=3\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
b) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{10x}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{5}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow4x-10x+10x=5+5-8\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
\(S=\left\{\dfrac{1}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Kết quả (x+1/2)^2 =
A. x^2+2x+1/4
B. x^2-x+1/4
C. x^2+x+1/4
D. x^2+x+0,2-1/3x)(0,21/2
2. Kết quả (x^2+2y)^2 bằng
A. 1/4+4y^2
B. 1/4+4y+4y^2
C. 1/4+2y+4y^2
D. 1/4+2y+2y^2
3. Kết quả phép tính (1/2x-0,5)^2 là
A. 1/2x^2-1/2x+0,25
B. 1/4x^2-0,25
C. 1/2x^2-0,25x+0,5
D. 1/4x^2-0,5x+0,25
4. Kết quả (0,2-1/3x)(0,2+1/3x)
A. 1/2x^2-1/2x+0,25
B. 1/4x^2-0,25
C. 1/2x^2-0,5x+2,5
D. Tất cả đều sai
5. Viết dưới dạng bình phương tổng x^2+2x+1 là:
A. (x+2)^2
B. (x+1)^2
C. (2x+1)^2
D. Tất cả đều sai
6. Kết quả (100a+5)^2 bằng
A. 100a^2+100a+25
B. 100a+100a+25
C. 100a^2-100a+25
D. 100a-100a+25
7. Kết quả thực hiện phép tính (2x-1/3)^2
A. 8x^3-1/27
B. 8x^2-2x^2+2/3x-1/27
8. Kết quả (1/2x-3)^2 = \(\frac{1}{4}x^2-3x+9\)
9. Với x=6 giá trị của đa thứcx^3+12x^2+48x+64 là
A. 900
B. 1000
C. 3000
D. Khác
10. Khi phân tích đa thức x^2-x kết quả là
A. x^2-x=x+1
B. x^2-x=x(x-1)
C. x^2-x=x
D. x^2-x=x^2(x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
\(2+\frac{x}{5}-0,5x=1-\frac{2x}{4}+0,25\)
\(2+\frac{x}{5}-\frac{1}{2}x=1+0,25-\frac{x}{2}\)
\(2+\frac{x}{5}-\frac{x}{2}=1,25-\frac{x}{2}\)
\(\frac{x}{5}-\frac{x}{2}+\frac{x}{2}=-2+1,25\)
\(\frac{2x-5x+5x}{10}=0,75\)
\(\frac{2x}{5}=0,75\)
\(2x=0,75.5\)
\(2x=3,75\)
\(x=\frac{3,75}{2}=\frac{15}{8}\)
Vậy \(x=\frac{15}{8}\)