2+4+8+....+2*x=210
Tim x
Những câu hỏi liên quan
2 phần ( x + 2 ). ( x + 4 ) + 4 phần ( x + 4 ) .( x + 8 ) + 6 phần( x + 8 ) .( x + 14 ) = x phần( x + 2 ) ( x + 14 ) với ( x ∉ { − 2 , − 4 , − 8 , − 14 } ) .
2. Tìm x biết:a)2(x+2)(x+4)dfrac{2}{left(x+2right)left(x+4right)} + 4(x+4)(x+8)dfrac{4}{left(x+4right)left(x+8right)} + 6(x+8)(x+14)dfrac{6}{left(x+8right)left(x+14right)} x(x+2)(x+14)dfrac{x}{left(x+2right)left(x+14right)}b)x2023dfrac{x}{2023} + x+12022dfrac{x+1}{2022}+ x+22021dfrac{x+2}{2021} +...+ x+20221dfrac{x+2022}{1} + 2023 0.Gíup mình giải 2 bài này với!Cảm ơn các bạn rất nhiều!!!
Đọc tiếp
2. Tìm x biết:
a)2(x+2)(x+4)\dfrac{2}{\left(x+2\right)\left(x+4\right)} + 4(x+4)(x+8)\dfrac{4}{\left(x+4\right)\left(x+8\right)} + 6(x+8)(x+14)\dfrac{6}{\left(x+8\right)\left(x+14\right)} = x(x+2)(x+14)\dfrac{x}{\left(x+2\right)\left(x+14\right)}
b)x2023\dfrac{x}{2023} + x+12022\dfrac{x+1}{2022} x+22021\dfrac{x+2}{2021} +...+ x+20221\dfrac{x+2022}{1} + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
Tinh gia tri cua phan so: A= 2 x 3 + 2 x 4 x 8 + 4 x 8 x 16 + 8 x 16 x 32/ 3x 4 + 2 x 6 x 8 + 4 x 12 x 16 + 8 x 24 x 32
Tìm x, biết:
2/(x+2).(x+4) + 4/(x+4).(x+8) + 8/(x+8).(x+16) = x/(x+2).(x+14)
Tính nhẩm
8 x 2 =
8 x 4 =
8 x 6 =
8 x 7 =
2 x 8 =
4 x 8 =
6 x 8 =
7 x 8 =
8 x 2 = 16
8 x 4 = 32
8 x 6 = 48
8 x 7 = 56
2 x 8 = 16
4 x 8 = 32
6 x 8 = 48
7 x 8 = 56
( có thể nhận xét như sau: Trong phép nhân khi đổi vị trí các thừa số thì tích không đổi)
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c) A= x(x+2)(x+4)(x+6)+8
A= x(x+6)(x+2)(x+4)+8
A= (x2+6x)(x2+6x+8)+8
Gọi x2+6x = a
A= a(a+8)+8
A= a2+8a+8= (a-4)2-8
A= (a-4)2-8 = (x2+6x+4)2-8\(\ge\)-8
Dấu bằng khi x2+6x+4= 0\(\Leftrightarrow\)x= -3-\(\sqrt{5}\)
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Phân tích đa thức thành nhân tử:
1.45+x^3-5*x^2-9*x
2.x^4-2*x^3-2*x^2-2*x+3
3.x^4-5*x^2+4
4.x^4+64
5.x^5+x^4+1
6.(x^2+2*x)*(x^2+2*x+4)+3
7.(x^3+4*x+8)^2+3*x*(x^2+4*x+8)+2*x^2
8. x^3*(x^2-7)^2-36*x
9.x^5+x+1
10. x^8+x^4+1
11. x^5-x^4-x^3-x^2-x-2
12. x^9-x^7-x^6-x^5+x^4+x^3+x^2-1
13. (x^2-x)^2-12*(x^2-x)+24
1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)
\(=\left(9-x^2\right)\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)
3, \(x^4-5x^2+4\)
Đặt \(x^2=t\left(t\ge0\right)\)ta có :
\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)
\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
`Answer:`
1. `45+x^3-5x^2-9x`
`=x^3+3x^2-8x^2-24x+15x+45x`
`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`
`=(x+3).(x^2-8x+15)`
`=(x+3).(x^2-5x-3x+15)`
`=(x-3).(x-5).(x-3)`
2. `x^4-2x^3-2x^2-2x-3`
`=x^4+x^3-3x^3+x^2+x-3x-3`
`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`
`=(x+1).(x^3-3x^2+x-3)`
`=(x+1).[x^3 .(x-3).(x-3)]`
`=(x+1).(x-3).(x^2+1)`
3. `x^4-5x^2+4`
`=x^4-x^2-4x^2+4`
`=x^2 .(x^2-1)-4.(x^2-1)`
`=(x^2-1).(x^2-4)`
`=(x-1).(x+1).(x-2).(x+2)`
4. `x^4+64`
`=x^4+16x^2+64-16x^2`
`=(x^2+8)^2-16x^2`
`=(x^2+8-4x).(x^2+8+4x)`
5. `x^5+x^4+1`
`=x^5+x^4+x^3-x^3+1`
`=x^3 .(x^2+x+1)-(x^3-1)`
`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`
`=(x^2+x+1).(x^3-x+1)`
6. `(x^2+2x).(x^2+2x+4)+3`
`=(x^2+2x)^2+4.(x^2+2x)+3`
`=(x^2+2x)^2+x^2+2x+3.(x^2+2x)+3`
`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`
`=(x^2+2x+1).(x^2+2x+3)`
`=(x+1)^2 .(x^2+2x+3)`
7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`
`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`
`=x^6+8x^4+19x^3+30x^2+88x+64`
8. `x^3 .(x^2-7)^2-36x`
`=x[x^2.(x^2-7)^2-36]`
`=x[(x^3-7x)^2-6^2]`
`=x.(x^3-7x-6).(x^3-7x+6)`
`=x.(x^3-6x-x-6).(x^3-x-6x+6)`
`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`
`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`
`=x.(x+1).(x-1).(x^2-3x+2x-6).(x^2+3x-2x-6)`
`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`
`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`
9. `x^5+x+1`
`=x^5-x^2+x^2+x+1`
`=x^2 .(x^3-1)+(x^2+x+1)`
`=x^2 .(x-1).(x^2+x+1)+(x^2+x+1)`
`=(x^2+x+1).(x^3-x^2+1)`
10. `x^8+x^4+1`
`=[(x^4)^2+2x^4+1]-x^4`
`=(x^4+1)^2-(x^2)^2`
`=(x^4-x^2+1).(x^4+x^2+1)`
`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`
`=[(x^2+1)^2-x^2].(x^4-x^2+1)`
`=(x^2-x+1).(x^2+x+1).(x^4-x^2+1)
11. ` x^5-x^4-x^3-x^2-x-2`
`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`
`=x^4 .(x-2)+x^3 ,(x-2)+x^2 .(x-2)+x.(x-2)+(x-2)`
`=(x-2).(x^4+x^3+x^2+x+1)`
12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`
`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`
`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`
`=(x^2-1).(x^7-x^4-x^3+1)`
`=(x-1)(x+1)(x^3-1)(x^4-1)`
`=(x-1)(x+1)(x^2+x+1)(x-1)(x^2-1)(x^2+1)`
`=(x-1)^2 .(x+1)(x^2+x+1)(x-1)(x+1)(x^2+1)`
`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`
13. `(x^2-x)^2-12(x^2-x)+24`
`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`
`=(x^2-x+6)^2-12`
`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`
Kết quả của phép tính (x - 2)(x - 4)(x−2)(x−4) là:
x^2 - 2x + 8x2−2x+8
x^2 +2x - 8x2+2x−8
x^2- 6x + 8x2−6x+8
x^2 + 6x -8x2+6x−8
Phân tích đa thức thành nhân tử.
1)x^4+2x^3-4x-4
2)(x+2)(x+4)(x+6)(x+8)+16
3)(x^2+x).(x^2+x+1)-6
4)(x^2+4x+8)^2+3x(x^2+4x+8)
ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)
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Tính
8 x 1 = ..... 8 x 2 = ..... 8 x 3 = ..... 8 x 4 = .....
1 x 8 = ..... 2 x 8 = ..... 3 x 8 = ..... 4 x 8 = .....
8 x 5 = ..... 8 x 6 = ..... 8 x 7 = ..... 8 x 8 = .....
5 x 8 = ..... 6 x 8 = ..... 7 x 8 = ..... 8 x 9 = .....
8 x 1 = 8 8 x 2 = 16 8 x 3 = 24 8 x 4 = 32
1 x 8 = 8 2 x 8 = 16 3 x 8 = 24 4 x 8 = 32
8 x 5 = 40 8 x 6 = 48 8 x 7 = 56 8 x 8 = 64
5 x 8 = 40 6 x 8 = 48 7 x 8 = 56 8 x 9 = 72
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