|3/4x-3/4|-3/4=|-3/4|
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
24 tháng 11 2021
\(a,=16x^8+8x^6\\ b,=4x^4-6x^5-4x^3\\ c,=15x^6+9x^3y-10x^3y-6y^2\\ =15x^6-x^3y-6y^2\\ d,=2a^4-a^3b+6a^2b-3ab^2-3ab^2+b^3\\ =2a^4-a^3b+6a^2b-6ab^2+b^3\)
DH
4 tháng 7 2017
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow10x-12x-12x=-16+11+16-15\)
\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x=182-36\)
\(\Rightarrow-73x=146\Rightarrow x=-2\)
Chúc bạn học tốt!!!
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
20 tháng 4 2022
`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
12 tháng 7 2018
Tìm GTNN của biểu thức :
\(x^2+2x+4\)
Đặt A = \(x^2+2x+4\)
\(\Leftrightarrow A=\left(x^2+2.x.1+1\right)+3\)
\(\Leftrightarrow A=\left(x+1\right)^2+3\)
Ta luôn có : \(\left(x+1\right)^2\ge0\forall x\)
Suy ra : \(\left(x+1\right)^2+3\ge3\forall x\)
Hay A\(\ge3\) với mọi x
Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)
Nên : \(A_{min}=3khix=-1\)
C
3
13 tháng 1 2022
320-4x+43=352
4x+64=320-352
4x+64=-32
4x=-32-64
4x=-96
x=-96:4
x=-24
2 tháng 1 2023
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
15 tháng 1 2017
a) \(^{x^3}\) - 7x+6=0
\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0
\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0
\(\Leftrightarrow\) x\(\left(x^2-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!
\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}-\frac{3}{4}=\text{|}-\frac{3}{4}\text{|}\)
<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}-\frac{3}{4}=\frac{3}{4}\text{ }\)
<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}=\frac{3}{4}+\frac{3}{4}\text{ }\)
<=>\(\text{|}\frac{3}{4}x-\frac{3}{4}\text{|}=\frac{3}{2}\)
<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{cases}}\)
<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x=\frac{3}{2}+\frac{3}{4}\\\frac{3}{4}x=-\frac{3}{2}+\frac{3}{4}\end{cases}}\)
<=>\(\orbr{\begin{cases}\text{ }\frac{3}{4}x=\frac{9}{4}\\\frac{3}{4}x=-\frac{3}{4}\end{cases}}\)
<=>\(\orbr{\begin{cases}\text{ }x=\frac{9}{4}:\frac{3}{4}=3\\x=-\frac{3}{4}:\frac{3}{4}=-1\end{cases}}\)
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-3=\frac{3}{2}\\\frac{3}{4}x-3=\frac{-3}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x=\frac{9}{4}\\\frac{3}{4}x=\frac{-3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)