1. (x+2).(x+4).(x+6).(x+8) +16 =0
2. (x+1).(x+2).(x+3).(x+4) -24 =0
3.(x-1).(x-3).(x-5).(x-7) -20 =0
giải phương trình hộ mik vs mn ơi huhu
giải phương trình sau
1/ 2x( x+3) - 6 (x-3) =0
2/ 2x^2( 2x+3) +(2x+3) =0
3/ (x-2) (x+1) -(x-2) 4x =0
4/ 2x ( x-5) -3x +15=0
5/ 3x(x+4) -2x-8 =0
6/ x^2 (2x-6) + 2x -6 =0
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
phân tích đa thức thành nhân tử
1.(x+2)(x+3)(x+4)(x+5)-24
2.(x-1)(x-3)(x-5)(x-7)-20
3.x(x-1)(x-1)(x+2)-3
4.(x-7)(x-5)(x-4)(x-2)-72
5.(x^2+6x+8)(x^2-8x+15)-24
6.(x^2-6x+5)(x^2-10x+21)-20
7.(x^2+5x+6)(x^2-15x+56)-144
#Mn làm giúp mik vs#
#thank kiu#
phân tích đa thức thành nhân tử
1.(x+2)(x+3)(x+4)(x+5)-24
2.(x-1)(x-3)(x-5)(x-7)-20
3.x(x-1)(x-1)(x+2)-3
4.(x-7)(x-5)(x-4)(x-2)-72
5.(x^2+6x+8)(x^2-8x+15)-24
6.(x^2-6x+5)(x^2-10x+21)-20
7.(x^2+5x+6)(x^2-15x+56)-144
#Mn làm giúp mik vs#
#thank kiu#
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x=t\)
\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)
\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)
Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)
\(\Delta=7^2-4.11=5\)
\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)
2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x=t\)
\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)
\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)
Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)
\(\Delta=\left(-8\right)^2-4.11=20\)
\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)
Nhầm, lộn tìm GTNN
a) \(\left(t+11\right)^2-25=\left(x^2+7x-16\right)\left(x^2+7x+36\right)\)
b) \(\left(x^2-8x+5\right)\left(x^2-8x+16\right)\)
\(=\left(x^2-8x+5\right)\left(x-4\right)^2\)
phân tích đa thức thành nhân tử
1.(x+2)(x+3)(x+4)(x+5)-24
2.(x-1)(x-3)(x-5)(x-7)-20
3.x(x-1)(x-1)(x+2)-3
4.(x-7)(x-5)(x-4)(x-2)-72
5.(x^2+6x+8)(x^2-8x+15)-24
6.(x^2-6x+5)(x^2-10x+21)-20
7.(x^2+5x+6)(x^2-15x+56)-144
#Mn làm giúp mik vs#
#thank kiu#
phân tích đa thức thành nhân tử
1.(x+2)(x+3)(x+4)(x+5)-24
2.(x-1)(x-3)(x-5)(x-7)-20
3.x(x-1)(x-1)(x+2)-3
4.(x-7)(x-5)(x-4)(x-2)-72
5.(x^2+6x+8)(x^2-8x+15)-24
6.(x^2-6x+5)(x^2-10x+21)-20
7.(x^2+5x+6)(x^2-15x+56)-144
#Mn làm giúp mik vs#
#thank kiu#
1/ 10(X-7)-8(X+5)=6(-5)+24
2/ 8(X-|-7|)-6(X-2)=|-8|.6-50
3/ 2(4X-8)-7(3+X)=|-4|(3-2)
4/ 12(X-4)=6(x-2)-16(X+3)=7|-4|
5/ 4(X-5)-7(5-X)+10(5-X)=-3
bn nào giải giúp mik vs nha:)))
1/ 10(X-7)-8(X+5)=6(-5)+24
10x - 70 - 8x - 40 = -30 +24
2x - 110 = -6
2x = 104
x=52
2/ 8(X-|-7|)-6(X-2)=|-8|.6-50
8(x - 7) - 6(x-2) = 8.6 - 50
8x - 56 - 6x +12 =48 -50
2x - 44 = -2
2x = 42
x=21
3/ 2(4X-8)-7(3+X)=|-4|(3-2)
8x-16 - 21 - 7x = 4.1
x-37=4
x=41
4/ 12(X-4)=6(x-2)-16(X+3)=7|-4|
12x - 48 = 6x - 12 - 16x -48 =7.4
12x - 48 = 28
12x=76
x=19/3
5/ 4(X-5)-7(5-X)+10(5-X)=-3
4x - 20 -35 +7x + 50 -10x = -3
x - 5 = -3
x = -2
Chúc bạn học tốt!
tìm x,bt:
a,16-18/x/=-8-14/x/
b,5./x/+7.(2/x/-3)=4.(/x/-4)+25
c,4/3x+2/-16=20
d,8/x-1/-4=12+6/x-1/
e,15-3/4x+2/=3
i,18-4.(6-2/x/)=3.(4/x/+5+29
g,27-5/x-10/=4/x-10/-18
h,5/10+3x/-21-7-2/15+3x/
mn ơi,giúp mk với T^T
Giair phương trình
\(8\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=\left(x+4\right)^2+4\left(x+\dfrac{1}{x}\right)^2\left(x^2+\dfrac{1}{x^2}\right).\)
giúp mik vs mn ơi
Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?
1,Giải các phương trình sau:
a,(x-4)^2-25 = 0
b,(x-3)^2-(x+1)^2 = 0
c,(x^2-4)(2x-3) = (x^2 - 4 )(x-1)
d,(3x-7)^2 - 4( x+1)^2 = 0
Giải giúp em với ạ :3
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a) Ta có: 4x-20=0
hay x=5
Vậy: S={5}
b) Ta có:
hay x=-4