y x 4 +2+4+6+8+...+18=256
y?
Cho y=(x-1):2. Cmr: (x+2y)(x²+4y²)(x⁴+16y⁴)=x8-256y8
từ giả thiết suy ra x-2y =1
=> (x+2y)(x2+4y2)(x4+16y4)=(x - 2y ) (x+2y)(x2+4y2)(x4+16y4)=
sd hằng đẳngthức để làm tiếp nhé
(2 điểm) Tính tổng
a. x = 1+ 2+ 3+ 4 + 5 + 6 + 7 + 8 + 9 +10
b. y = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
a. x = 1+ 2+ 3+ 4 + 5 + 6 + 7 + 8 + 9 +10
= (1 +10) + (2+9) + ( 3 +8 ) + (4 + 7 ) + (5 + 6)
= 11 +11 + 11+ 11 +11
= 11 x 5
= 55
b. y = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
= (2+20) + (4 + 18) + (6 + 16) +(8 + 14) + (10 +12)
= 22 + 22 + 22 +22 + 22
= 22 x 5
= 110
tìm các số nguyên x,y,z a) -4/8 = x/-10 = -7/y = z/-24 b) -3/6 = x/-2 = -18/y = -z/24
a: \(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\)
=>\(\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\left\{{}\begin{matrix}x=\left(-10\right)\cdot\dfrac{\left(-1\right)}{2}=5\\y=\dfrac{-7\cdot2}{-1}=14\\z=\dfrac{-24\cdot\left(-1\right)}{2}=\dfrac{24}{2}=12\end{matrix}\right.\)
b: \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
=>\(\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\dfrac{x}{2}=\dfrac{18}{y}=\dfrac{z}{24}=\dfrac{1}{2}\)
=>\(x=2\cdot\dfrac{1}{2}=1;y=18\cdot\dfrac{2}{1}=36;z=\dfrac{24}{2}=12\)
Tìm số nguyên x, y biết:
a, \(\dfrac{-1}{5}\)≤ \(\dfrac{x}{8}\)≤ \(\dfrac{1}{4}\)
b, \(\dfrac{4}{x-6}\)= \(\dfrac{y}{24}\)= \(\dfrac{-12}{18}\)
c, \(\dfrac{x+46}{20}\)=x \(\dfrac{2}{5}\)
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
a) 16 - 18/x| = -8 -14|x|
b) 5 x|+7.(2/x|-3) =4.(x-4)+25
e) 15-3|4x +2| = 3 8) 18 -4.(6-2|x|) = - 3.(4|x|+5)+29
8) 27-5/x -10 = 4|x-10| - 18
h) 510 + 3x|-21 = 7- 2|15+3x|
c) 43x +2|-16= 20
d) 8|x - 1|-4 = 12+6|x-1|
giúp em với
Tìm các số nguyên x,y biết:
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
b) \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
d) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
f) \(y\dfrac{5}{y}=\dfrac{86}{y}\) ( \(x\dfrac{2}{5};y\dfrac{5}{y}\) là các hỗn số)
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)
Đúng ghi Đ, sai ghi S:
a) 32 : 4 : 2 = 8 : 2 = 4
32 : 4 : 2 = 32 : 2 = 16
b) 18 : 2 x 3 = 18 : 6 = 3
18 : 2 x 3 = 9 x 3 = 18
a) 32 : 4 : 2 = 8 : 2 = 4
32 : 4 : 2 = 32 : 2 = 16
b) 18 : 2 x 3 = 18 : 6 = 3
18 : 2 x 3 = 9 x 3 = 18
Đúng ghi Đ sai ghi S:
a,5 x 7/18 + 2/9 = 5 x 11/18=55/18
b,3/4 x (2/9+5/6) = 3/4 x 2/9 +3/4 x 5/6=1/6 + 5/8=19/24
Bài 3: Rút gọn biểu thức (Dùng hằng đẳng thức)
1, (x+y)\(^2\)-(x-y)\(^2\)
2, (x+y)\(^3\)-(x-y)\(^3\)-2y\(^3\)
3,(x+y)\(^2\)-2(x+y)(x-y)+(x-y)\(^2\)
4,(2x+3)\(^2\)-2(2x+3)(2x+5)+(2x+5)\(^2\)
5, 9\(^8\). 2\(^8\)-(18\(^4\)+1)(18\(^4\)-1)
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1