(2x-3)^4=(2x-3)^2
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16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
TA
21 tháng 7 2021
1) `2x(3x-1)-(2x+1)(x-3)`
`=6x^2-2x-2x^2+6x-x+3`
`=4x^2+3x+3`
2) `3(x^2-3x)-(4x+2)(x-1)`
`=3x^2-9x-4x^2+4x-2x+2`
`=-x^2-7x+2`
3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`
`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`
`=3x^2-15x-x^2+4x-4-4x^2+9`
`=-2x^2-11x+5`
4) `(2x-3)^2+(2x-1)(x+4)`
`=4x^2-12x+9+2x^2+8x-x-4`
`=6x^2-5x+5`
2 tháng 11 2019
+) Lỗi nhỏ: Sai ở chỗ: \(\left|x-2+4-3x\right|=\left|-2x-2\right|\)
+) Lỗi lớn: Dấu bằng xảy ra: \(\hept{\begin{cases}\left(x-2\right)\left(4-3x\right)\ge0\\\left(-2x+2\right)\left(2x-3\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{3}{2}\le x\le1\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le1\)( làm tắt )
Nhưng mà thử vào chọn x= 1=> A = 3 > 1. Nên bài này sai.
Làm lại nhé!
A = | x - 2 | + | 2 x - 3 | + | 3 x - 4 |
= | x - 2 | + | 2 x - 3 | + 3 | x - 4/3 |
= | x -2 | + | x - 4/3 | + | 2x -3 | +2 | x - 4/3 |
= ( | 2 - x | + | x - 4/3 | ) + ( | 3 - 2x | + | 2x - 8/3 | )
\(\ge\)| 2 -x + x - 4/3 | + | 3 - 2x + 2x -8/3 |
= 2/3 + 1/3 = 1
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(2-x\right)\left(x-\frac{4}{3}\right)\ge0\\\left(3-2x\right)\left(2x-\frac{8}{3}\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{4}{3}\le x\le\frac{3}{2}\end{cases}}\Leftrightarrow\frac{4}{3}\le x\le\frac{3}{2}\)
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
30 tháng 1 2021
ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-1;\dfrac{-3}{2};-2\right\}\)
Ta có: \(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)
\(\Leftrightarrow\dfrac{4\left(2x+3\right)}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{2\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2}{\left(2x+2\right)\left(2x+4\right)}-\dfrac{3\left(2x+4\right)}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{8x+12-4x-2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2-6x-12}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{4x+10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\left(4x+10\right)\left(\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{1}{\left(2x+2\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow2\left(2x+5\right)\left(\dfrac{\left(2x+2\right)\left(2x+4\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}+\dfrac{\left(2x+1\right)\left(2x+3\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(4x^2+8x+4x+8+4x^2+6x+2x+6\right)=0\)(Vì \(\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left(2x+5\right)\left(8x^2+20x+14\right)=0\)
mà \(8x^2+20x+14>0\forall x\)
nên 2x+5=0
\(\Leftrightarrow2x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
Vậy: \(S=\left\{-\dfrac{5}{2}\right\}\)
13 tháng 4 2020
\(24x-4\left(2x-\frac{3}{4}\right)-4\left(3+\frac{2x}{2}\right)=36-3\left(x-\frac{3}{2}\right)-3\left(3-\frac{2x}{3}\right)\)
Đề như này đúng không bạn
8 tháng 4 2023
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
23 tháng 4 2023
cho A=1/101+1/102+1/103+...+1/199+1/200 chứng minh 1/2 <A<1
10 tháng 9 2023
a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2
=>25-4x^2+2x^2+6x=4-2x^2
=>6x+25=4
=>6x=-21
=>x=-7/2
b: (3x-2)(-2x)+5x^2=-x(x-3)
=>-6x^2+4x+5x^2=-x^2+3x
=>4x=3x
=>x=0
c: =>7-(4x^2-9)=x^2+8x+16
=>7-4x^2+9-x^2-8x-16=0
=>-5x^2-8x=0
=>5x^2+8x=0
=>x(5x+8)=0
=>x=0 hoặc x=-8/5
=>(2x-3)^4-(2x-3)^2=0
=>(2x-3)^2.[(2x-3)^2-1]=0
TH1:(2x-3)^2=0=>x=3/2
TH2:(2x-3)^2-1=0=>x=2;1