a)|1/3.x-1/4|-1/6=1/2
b)|4/7.x+3/4-1/3=1/2
c)|2/3x-25%|+1/4+1 và 3/4
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Những câu hỏi liên quan
T
13 tháng 8 2023
a) (3x - 2)2 - (1 + 5x)2
= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)
= (-2x - 3)(8x - 1)
b) (3x + 4)(3x - 4) - (5 - x)2
= (3x)2 - 42 - (25 - 10x + x2)
= 9x2 - 16 - 25 + 10x - x2
= 8x2 + 10x - 41
c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)
\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)
\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)
\(=4x+25\)
13 tháng 8 2023
a: =9x^2-12x+4-25x^2-10x-1
=-16x^2-22x+3
b: =9x^2-16-x^2+10x-25
=8x^2+10x-41
c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)
13 tháng 8 2023
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
26 tháng 5 2017
1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
ML
26 tháng 5 2017
Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi
.
, ta có: 
.![(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]](https://s0.wp.com/latex.php?latex=%283-x%29%282%2Bx%29+%5Cleq+%5Cdfrac%7B25%7D%7B4%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-2%3B+3%5D&bg=ffffff&fg=4e4e4e&s=0 "(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]")
.![(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]](https://s0.wp.com/latex.php?latex=%282-x%29.%281%2B2x%29+%5Cleq+%5Cdfrac%7B25%7D%7B8%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-%5Cdfrac%7B1%7D%7B2%7D%3B+2%5D&bg=ffffff&fg=4e4e4e&s=0 "(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]")
LP
17 tháng 8 2021
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
a: \(\Leftrightarrow\left|x\cdot\dfrac{1}{3}-\dfrac{1}{4}\right|=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{2}{3}\\x\cdot\dfrac{1}{3}-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{4}{7}+\dfrac{3}{4}\right|=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{4}{7}+\dfrac{3}{4}=\dfrac{1}{6}\\x\cdot\dfrac{4}{7}+\dfrac{3}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-49}{48}\\x=-\dfrac{77}{48}\end{matrix}\right.\)