1) CMR: 543-54 khong la so chinh phuong
2) Tim x:
2(x-2).(x+3)-x2+4=0
3) Rut gon
a)2(x+1)2-3(x-1)2+(x+2).(5-x)
b)(3x-1)3+(3x-1)3-6x2+9
4) A= (x-5).(x+2)+3.(x-2).(x+2)-(3x-1)2+5x2
a) rut gon A
b) tinh a khi x =1/2
1) CMR: 543-54 khong la so chinh phuong
2) Tim x:
2(x-2).(x+3)-x2+4=0
3) Rut gon
a)2(x+1)2-3(x-1)2+(x+2).(5-x)
b)(3x-1)3+(3x-1)3-6x2+9
4) A= (x-5).(x+2)+3.(x-2).(x+2)-(3x-1)2+5x2
a) rut gon A
b) tinh a khi x =1/2
B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy
cho M=((x+2/3x)+(2/x+1)-3):(2-4x/x+1)-(3x-3x^2+1/3x)
a. rut gon M
b. tim x sao cho M<1/3
c. tim xϵZ de MϵZ
giup mk vs, mk dang can gap a
tinh hop li:
12-12+11+10-9+8-7+5-4+3+2-1
tim so nguyen x :
3x+27=9
2x+12=3(x-7)
2x^2-1=49
-|9-x|-5=12
cho:A=(-a-b+c)-(-a-b-c)
a)rut gon A
b)tinh gia tri cua A khi a=1;b= -1;c= -2
Bài 1 : 12 - 12 + 11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 + 2 - 1
= ( 12 - 12 ) + ( 11 - 1 ) + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + ( 3 + 2 )
= 0 + 10 + 1 + 1 + 1 + 5
= 18
Bài 2 :
3x + 27 = 9
3x = 9 - 27
3x = - 18
x = - 6
2x + 12 = 3( x - 7 )
2x + 12 = 3x - 21
3x - 2x = 12 + 21
x = 33
2x2 - 1 = 49
2x2 = 49 + 1
2x2 = 50
x2 = 50 : 2
x2 = 25
=> x = 5 hoặc x = - 5
- | 9 - x | - 5 = 12
- | 9 - x | = 12 + 5
- | 9 - x | = 17
TH1 : 9 - x >= 0 <=> x <= 9
=> - ( 9 - x ) = 17
=> x = 26 ( loại )
TH2 : 9 - x < 0 <=> x > 9
=> - ( 9 - x ) = -17
=> x = - 8 ( loại )
=> ko có giá trị nào thõa mãn
Bài 3 a,: A = ( - a - b + c ) - ( - a - b - c )
= - a - b + c + a + b + c
= 2c
b, thay c = - 2 vào biểu thức A = 2c
Ta được : A = 2 x ( -2 ) = - 4
a- a2(a+1)+2a(a+1) chia het cho 6 vs a la so nguyen
b- x2+2x+2>0 vs moi x
b) Ta có:
\(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\)
Suy ra đpcm.
1. (2x - 1)(3x + 2)(3 - x)
= (6x2 + 4x - 3x - 2)(3 - x)
= 18x2 + 3x - 6 - 6x3 - x2 + 2x
= 6x3 + 17x2 +5x - 6
3. x2 - 4x - 5 = x2 + x - 5x - 5 = x(x + 1) - 5(x + 1) = (x - 5)(x + 1)
x2 - 25 - 2xy + y2 = (x2 - 2xy + y2) - 25 = (x - y)2 - 52 = (x - y + 5)(x - y - 5)
x4 + x2 + 1 = x4 + 2x2 + 1 - x2 = (x2 + 1)2 - x2 = (x2 + 1 - x)(x2 + 1 + x)
4.
a. a2(a + 1) + 2a(a + 1) = (a + 1)(a2 + 2a) = a(a + 1)(a + 2)
Vì a(a + 1)(a + 2) là tích của 3 số nguyên liên tiếp => a(a + 1)(a + 2) chia hết cho 2 và 3
Mà ƯCLN(2,3) = 1 => a(a + 1)(a + 2) chia hết cho 6 (đpcm)
b. x2 + 2x + 2 = x2 + 2x + 1 + 1 = (x + 1)2 + 1
Vì (x + 1)2 \(\ge\)0 với mọi x => (x + 1)2 + 1 > 0 với mọi x (đpcm)
a, rut gon A
b, tim x de a<-1
c, tim cac gia tri nguyen cua x de A co gia tri nguyen
cho bthuc B = \(\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x-2}\right)chia\left(x-2+\frac{16-x^2}{x+2}\right)\)rut gon B tính b khi /x/ = 1/2tim x de b=2tim x \(\in\) z de b \(\in\) zBài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
Tìm x
a)(2x+1)2-4(x+2)2 =9
b)(3x-1)2 +2(x+3)2 +11(x+1)(1-x)=6
c)(x+1)3 - x2 (x+3)=2
d)(x-2)3 -x(x+1)(x-1)+6x2 =5
e)(x-3)(x2 +3x +9)-x(x+4)(x-4)=5
g)(x-2)3 -(x+5)(x2 -5x+25)+6x2 =11
Bài 1. Tìm x, biết
a) (x+4)2-x2(x+12)=16
c) (x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)=28
d) (x-2)3-(x+5)(x2-5x+25)-6x2=11
Bài 2. Rút gọn các biểu thức sau:
A = (x+1)3+(x-1)3
B = (x-3)3-(x+3)(x2-3x+9)+(3x-1)(3x+1)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
Tìm x
a)(2x+1)2-4(x+2)2 =9
b)(3x-1)2 +2(x+3)2 +11(x+1)(1-x)=6
c)(x+1)3 - x2 (x+3)=2
d)(x-2)3 -x(x+1)(x-1)+6x2 =5
e)(x-3)(x2 +3x +9)-x(x+4)(x-4)=5
g)(x-2)3 -(x+5)(x2 -5x+25)+6x2 =11
\(\left(2x+1\right)2-4\left(x+2\right)2=9\)
\(4x+2-8x-16=9\)
\(4x-8x=9+16-2\)
\(-4x=23\)
\(x=-\frac{23}{4}\)
a, \(\left(2x+1\right)2-4\left(x+2\right)2=9\)
\(\Leftrightarrow4x+2-8x-16=0\Leftrightarrow-4x-14=0\Leftrightarrow x=-\frac{7}{2}\)
b, \(\left(x+1\right)3-2x\left(x+3\right)=2\)
\(\Leftrightarrow3x+3-2x^2-6x=2\Leftrightarrow-3x+1-2x^2=0\)