x(5-12)+8(7-16)+12x-15(17-56)=0

​x.(-7)+8.(-9)+12x-15.(-39)=0

​x.(-7)+12x-72+585=0

​x(-7+12)+513=0

​x.5=-513

​x=-513/5

​

kết bn đi

\(B=x^{17}-12.x^{16}+12.x^{15}-12.x^{14}+...-12.x^2+12x-1\)

\(=11^{17}-\left(11+1\right)11^{16}+\left(11+1\right)11^{15}-\left(11+1\right)11^{14}+...-\left(11+1\right)11^2+\left(11+1\right)11-1\)

\(=11^{17}-11^{17}-11^{16}+11^{16}+11^{15}-11^{15}-11^{14}+...-11^3-11^2+11^2+11-1\)

\(=11-1=10\)

Vậy B = 10

B= x^17 - 12x^16 +12x^15 - 12x^14 +...-12x^2+12x-1

  = x^17-11x^16-x^16+11x^15 +x^15-11x^14-x^14+...-x^2+11x+x-1

  = (x-11)( x^16-x^15+x^14-...-x +1) 

  = 0 ( vì x=11)

2x^2 + x - 6

= 2x^2 + 4x - 3x - 6

= 2x(x + 2) - 3(x + 2)

= (2x - 3)(x + 2)

7x^2 + 50x + 7 

= 7x^2 + x + 49x + 7

= 7x(x + 7) + x + 7

= (7x + 1)(x + 7)

12x^2 + 7x - 12

15x^2 +  7x - 2

= 15x^2 - 3x + 10x - 2

= 3x(5x - 1) + 2(5x - 1) 

= (3x + 2)(5x - 1)

a^2 - 5a - 14

= a^2 + 2a - 7a - 14

= a(a + 2) - 7(a + 2)

= (a - 7)(a + 2)

2x^2 + 5x + 2

= 2x^2 + x + 4x + 2

= 2x(x + 2) + x + 2

= (2x + 1)(x + 2)

\(2x^2+x-6=2x^2+4x-3x-6\)

\(=2x\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2x-3\right)\)

\(7x^2+50x+7\)

\(=7x^2+x+49x+7\)

\(=x\left(7x+1\right)+7\left(7x+1\right)\)

\(=\left(7x+1\right)\left(x+7\right)\)

\(12x^2+7x-12\)

\(=12x^2+16x-9x-12\)

\(=4x\left(3x+4\right)-3\left(3x+4\right)\)

\(=\left(3x+4\right)\left(4x-3\right)\)

\(4p^2-36p+56\)

\(=4p^2+8p+28p+56\)

\(=4p\left(p+2\right)+28\left(p+2\right)\)

\(=\left(p+2\right)\left(4p+28\right)\)

a/ => x - 2 = -25 => x = -23

b/ => 15 - x = -4 => x = 19

c/ => 12x - 91 = 1/101

=> 12x = 9192/101

=> x = 766/101

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)

Thay x = 11 ta có:

\(A=11^{100}-12.11^{99}+12.11^{98}-...-12.11^3+12.11^2-12.11+12\)

\(=11^{100}-12\left(11^{99}-11^{98}+11^{97}-...+11^3-11^2+11\right)+12\)

Đặt \(B=11^{99}-11^{98}+...+11\)

\(\Rightarrow11B=11^{100}-11^{99}+...+11^2\)

\(\Rightarrow12B=11^{100}+11\)

\(\Rightarrow B=\dfrac{11^{100}+11}{12}\)

Từ đó, \(A=11^{100}-12.\dfrac{11^{100}+11}{12}+12\)

\(=11^{100}-11^{100}-11+12=1\)

Vậy A = 1

Ta có: \(x=11\Rightarrow x+1=12\)

Khi đó, ta được:

\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)

\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-\left(x+1\right)x^{97}+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+12\)

\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-x^{98}-x^{97}+...-x^4-x^3+x^3+x^2-x^2-x+12\)

\(=\left(x^{100}-x^{100}\right)-\left(x^{99}-x^{99}\right)+\left(x^{98}-x^{98}\right)-...-\left(x^3-x^3\right)+\left(x^2-x^2\right)-x+12\)

\(=0-x+12=0-11+12=-11+12=1\)

Vậy tại x=11 thì A=1

Câu 9:

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(9,x^3-2x^2-x+2=0\\ \Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

\(10,\) giống 9

\(11,x^2+4x-5=0\\ \Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(12,2x^2+4x+2=72\\ \Rightarrow2x^2+4x-70=0\\ \Rightarrow x^2+2x-35=0\\ \Rightarrow\left(x^2-5x\right)+\left(7x-35\right)=0\\ \Rightarrow x\left(x-5\right)+7\left(x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

\(13,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ \Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\\ \Rightarrow x^3-25x-x^3-8=17\\ \Rightarrow-25x=25\\ \Rightarrow x=-1\)

\(14,2x^3+5x^2-12x=0\\ \Rightarrow x\left(2x^2+5x-12\right)=0\\ \Rightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Rightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ \Rightarrow x\left(2x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

Vì 143 - 12x chia hết cho x mà 12x chia hết cho x nên 143 chia hết cho x

Khi phân tích 143 ra thừa số nguyên tố thi 143=11*13 =>143 chia hết cho 11 và 13

=> x=11 hoặc 13 mà x<12 => x=11

Mà 143 - 12x chia hết cho 1

=> x cũng có thể =1

Vậy x=1 hoặc 11

p làm như vậy các em lớp 6 khó hiểu lắm ạ

Đề hơi khó hiểu nhưng vẫn biết cách làm !!!

Bài giải 

a) +)Ta có :  4x - 7 =   12x +5

=> 4x - 12x = 5 + 7 

<=> -8x     = 12

<=> x =\(\frac{-12}{8}=\frac{-3}{2}\)

+)Ta có :  2x -1 = 6x + 5 

<=> 2x - 6x      = 5 + 1 

<=>  -4x            = 6

<=> x = \(\frac{-6}{4}=\frac{-3}{2}\)

=> đây là cặp phương trình tương đương .

b) +) 7.( x - 10 ) =12 

+) 14 . ( x - 10 ) = 24

<=> \(\frac{1}{2}.\left[14.\left(x-10\right)\right]=\frac{1}{2}.24\)

<=>7 . ( x - 10 ) = 12 

=> Đây là 2 phương trình tương đương .

c) +) \(\frac{4}{x+3}-3=\frac{4}{x+3}+x.\left(ĐK:x\ne-3\right)\)

<=> \(\left(\frac{4}{x+3}-\frac{4}{x+3}\right)-3=x\)

<=> 0 - 3 = x 

<=>x = 3 

+) Với x= -3 => x + 3 = 0

=> ko thỏa mãn 

=> ko xét tính tương đương