3x-4
rút gọn b=(3x+4)^2+(5-3x)^2+2(3x+4)(5-3x)
\(\left(3x+4\right)^2+\left(5-3x\right)^2+2.\left(3x+4\right)\left(5-3x\right)\\ =\left[\left(3x+4\right)+\left(5-3x\right)\right]^2\\ =\left(3x+4+5-3x\right)^2\\ =9^2=81\)
Kết quả rút gọn biểu thức (3x+2).(3x-2) là A) 3x^2+4 B)3x^2-4 C)9x^2+4. D)9x^2-4
\(\left(3x+2\right)\left(3x-2\right)=9x^2-4\)
-> chọn D
x(1-3x)(4-3x)-(x-4)(3x+5)
x(1-3x)(4-3x)-(x-4)(3x+5)=(x-3x2)(4-3x)-(x-4)(3x+5)
=(4(x-3x2)-3x(x-3x2))-(3x(x-4)+5(x-4))
= (4x-12x2-3x2+9x3)-(3x2-12x+5x-20)
= (9x3-15x2+4x)-(3x2-7x-20)
= 9x3-15x2+4x-3x2-7x-20
= 9x3-18x2+x-20
p(x)=-2x+12x^2+3x^4-3x^2-3
g(x)=3x^4+x^2-4x^2+1,5x^2-3x^4+2x+1
thu gọn
\(P\left(x\right)=3x^4+9x^2-2x-3\)
\(Q\left(x\right)=\left(3x^4-3x^4\right)+\left(x^2-4x^2+1.5x^2\right)+2x+1=-1.5x^2+2x+1\)
B) (2x+3)2-(5x-4) (5x+4)=(x+5)2-(3x-1) (7x+2)-(x2-x+1)
C) (1-3x)2-(x-2) (9x+1)=(3x-4) (3x+4)-9(x+3)2
D) (3x+4) (3x-4) - (2x+5)2=(x-5)2+(2x+1)2-(x2-2x)+(x-1)2 cần gắp
tính
-4x.2xy^2+3x^2.1/3y+(-5)xy.(1/5xy)
4/3x^4y^7-3x^4y^7
2/3x^3y^4+3x^3y^4
A) \(-4x2xy^2+3x^2.\frac{1}{3}y+\left(-5\right)xy.\frac{1}{5}xy=-8x^2y^2+x^2y+\left(-x^2y^2\right)=-9x^2y^2+x^2y\)
B) \(\frac{4}{3}x^4y^7-3x^4y^7=\frac{-5}{3}x^4y^7\)
C) \(\frac{2}{3}x^3y^4+3x^3y^4=3\frac{2}{3}x^3y^4\)
CHÚC BN HỌC TỐT!!!
Tính :
1) 4(3x+4)
2) 4/3(5/3x+6)
3)(4/3x+7).5+2
4)(3/7x-6).(-7)+3x
1, 4[3x + 4] = 12x + 16
2, 4/3[5/3x + 6] = 20/9x + 12
3, [4/3x + 7]. 5 + 2 = 20/3x + 35 + 2 = 20/3x + 37
4, [3/7x - 6][-7] + 3x = -3x + 42 + 3x = 42
đường thẳng đi qua A<0,4> và vuông góc với đường thẳng y bằng 1/3x-7/3 có phương trình là
A y bằng -3x+4 B y bằng -3x-4 C y bằng 3x+4 D y bằng -1/3x +4
PT của đường thẳng cần tìm có dạng: `(d): y=ax+b `
`(d)` vuông góc `(d') : y=1/3 x-7/3 <=> a. 1/3 = -1 <=> a=-3`
`=> y=-3x+b`
`A (0;4) \in (d) <=> 4=-3.0+b <=> b=4`
`=> y=-3a+4`.
C=1/3x-2 -4/3x+2 -3x-6/4-9x2
a) ĐKXĐ : \(\left\{{}\begin{matrix}3x-2\ne0\\3x+2\ne0\\4-9x^2\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{2}{3}\)
\(C=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{4.\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-4.\left(3x-2\right)+3x-6}{\left(3x-2\right).\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right).\left(3x+2\right)}\)
\(=\dfrac{-2}{3x+2}\)
b) Với \(x\inℤ\)
Ta có : \(C\inℤ\Leftrightarrow-2⋮3x+2\)
\(\Leftrightarrow3x+2\inƯ\left(-2\right)\)
\(\Leftrightarrow3x+2\in\left\{1;2;-1;-2\right\}\)
Lập bảng
3x + 2 | 1 | 2 | -2 | -1 |
x | \(-\dfrac{1}{3}\left(\text{loại}\right)\) | 0(tm) | \(-\dfrac{4}{3}\left(\text{loại}\right)\) | -1(tm) |
Vậy \(x\in\left\{0;-1\right\}\)
Ta có :
Lập bảng
3x + 2 | 1 | 2 | -2 | -1 |
x | 0(tm) | -1(tm) |
Vậy