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1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
\(x-\frac{6}{7}+x-\frac{7}{8}+x-\frac{8}{9}=x-\frac{9}{10}+x-\frac{10}{11}+x-\frac{11}{12}\)
\(x+x+x-x-x-x=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
\(0=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
X triệt tiêu hết ròi! Vậy đề bài yêu cầu tìm gì vậy. Nhưng mà...giá trị của 2 vế ko bằng nhau.
\(\Leftrightarrow\left(\frac{x+1}{7}-1\right)+\left(\frac{x+1}{8}-1\right)+\left(\frac{x+1}{9}-1\right)=\left(\frac{x+1}{10}-1\right)+\left(\frac{x+1}{11}-1\right)+\left(\frac{x+1}{12}-1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=0\)
\(\text{Vì}\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\ne\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\)\(\Rightarrow\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(\Leftrightarrow\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1\)\(+\frac{x-11}{12}+1\) ( cộng 2 vế với 3 )
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\right)\)
\(\Leftrightarrow x=-1\)
Tính nhanh mỗi biểu thức sau:
a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10
= 20 x 10 + 10
= 200 + 10
= 210
b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0
= A x 0
= 0
c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 : A
= 0
d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (30 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (37 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x A
= 0
e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= 0 : (1 x 2 x 3 x 4 x ... x 10)
= 0 : A
= 0
g, (m : 1 - m x 1) : (m x 2008 + m x 2008)
= (m - m) : (m x 2008 + m x 2008)
= 0 : (m x 2008 + m x 2008)
= 0 : A
= 0
h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)
= (2 + 4 + 6 + 8 + m x n) x (972 - 972)
= (2 + 4 + 6 + 8 + m x n) x 0
= A x 0
= 0
l, (1 + 2 + 3 + ... + 99) x (13 x 15 - 12 x 15 - 15)
= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))
= (1 + 2 + 3 + ... + 99) x (15 x 0)
= (1 + 2 + 3 + ... + 99) x 0
= A x 0
= 0
i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)
= 0 x : (2 + 4 + 6 +...+ 98)
= 0 x A
= 0
k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)
= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))
= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)
= (0 + 1 + 2 +...+ 97 + 99) x 0
= A x 0
= 0
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})`
Ta có:
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
= 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= (36 – 36) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
= 0
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= (27 – 27) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7
=0
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7 Nếu đúng thì k cho mình nhé bạn!
Ta có
(x+8)/10+(x+7)/11+(x+6)/12=(x+5)/13+(x+4)/14-1
<=>(x+8)/10+1+(x+7)/11+1+(x+6)/12+1=(x+5)/13+1+(x+4)/14-1+2( giải thích là cộng mỗi vế với 3)
<=>(x+18)/10+(x+18)/11+(x+18)/12=(x+18)/13+(x+18)/14
<=>(x+18)/10+(x+18)/11+(x+18)/12-(x+18)/13-(x+18)/14=0
<=>(x+18)(1/10+1/11+1/12-1/13-1/14)=0
Hai số nhân lại bằng không khi và chỉ khi một trong hai số bằng không hoặc cả hai số bằng không mà
1/10+1/11+1/12-1/13-1/14 khác không
<=>x+18=0
<=>x=-18
Nhớ k mik nhé
Chọn đáp án D