(x+2)(x+3)(17x2 -17+8)=(x+2)(x+3)(x2 -17x+33)
\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
=>\(17x^2-17x+8=x^2-17x+33\)
<=> \(16x^2-25=0\)
<=>\(\left(4x-5\right)\left(4x+5\right)=0\)
=> \(4x-5=0=>x=\dfrac{5}{4}\)
hoặc \(4x+5=0=>x=\dfrac{-5}{4}\)
(x+2)(x−3)(17x2−17x+8)=(x+2)(x−3)(x2−17x+33)
\(\Leftrightarrow\)(x+2)(x−3)(17x2−17x+8) - (x+2)(x−3)(x2−17x+33) = 0
\(\Leftrightarrow\)(x+2)(x−3).[(17x2−17x+8)-(x2−17x+33)] = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x+2 = 0}\\\text{x−3 = 0}\\\text{(17x^2−17x+8)-(x^2−17x+33) = 0}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\17x^2-17x+8-x^2+17x-33=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\16x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\\left(4x-5\right)\left(4x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x-5=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x=5\\4x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{5}{4}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
Vậy S = \(\left\{-2;\dfrac{-5}{4};\dfrac{5}{4};3\right\}\)
f(x) = x17-17x16-17x15-.....+17x2+17x+17. tính f(16)
x=16 => x+1 = 17
f(x) = x^17 - 17x^16 - 17x^15 - ... + 17x^2 + 17x + 17
= x^17 - (x+1)x^16 - (x+1)x^15 - ... + (x+1)x^2 + (x+1)x + 17
= x^17 - x^17 + x^16 - x^16 + x^15 + ... + x^3 + x^3 + x^2 + x^2 + x + 17
= 2x^3 + 2x^2 + x + 17 = 2.16^3 + 2.16^2 + 16 + 17 = 8737
Mình gõ trên điện thoại nên bạn cố nhìn nhé ^^
À đâu, mình giải sai do nhầm dấu rồi :)))
Sr bạn nhiều -))
f(x) = x17-17x16-17x15-.....+17x2+17x+17. tính f(16)
Nó bắt đầu chuyển từ trừ sang cộng bắt đầu từ số nào?
Đề bài thiếu dữ kiện thế
Tìm số nguyên x:
a) -2x -(x-17)=34 -(-x+25)
b) 17x -(16x-37)=2x+43
c) -2x - 3.(x-17)=34-2(-x+25)
d) 17x + 3.(-16-37)=2x + 43 -4x
e) {-3x + 2.[45 - x - 3(3x + 7)-2x] + 4x}=55
Giúp mik vs:33
a,-2x -(x-17)=34-(-x+25)
-2x-x+17=34+x-25
-3x+17=9+x
-3x-x=9-17
-4x=-8
-->4x=8
x=8:4
x=2
Vậy x=2
b,17-(16x-37)=2x+43
17-16x+37=2x+43
20-16x=2x+43
-16x-2x=43-20
-18x=23
x=23:(-18)
x=23/-18
Mà x là số nguyên nên --> x thuộc tập rỗng
c,-2x-3.(x-17)=34-2(-x+25)
-2x-3x+51=34-2.(-x)-25
-5x+51=9-(-2).x
-5x+(-2).x=9-51
-7x=-42
7x=42
x=42:7
x=6
Vậy x=6
Đa thức x^4+3x^3-17x^2+ax+bx4+3x3−17x2+ax+b chia hết cho đa thức x^2+5x-3x2+5x−3 thì giá trị của biểu thức a+ba+b là
a) ( 21/22 + 3/5 + -21/22 ) : 1 2/5 9/17 x 8/5 - 9/17 x 3/5 + 8/17
X - 3/10 = 7/15 x 3/15
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)