x+4x+7x+10x+........+58x=595
Tìm x
x+4x+7x+.......+58x=596
tìm x
Ta có : x + 4x + 7x + ...... + 58x = 596
<=> x(1 + 4 + 7 + ...... + 58) = 596
=> x 590 = 596
=> x = 596/590
x + 4x + 7x + ..... + 58x = 596
( x + 4x + 7x + ...+58x) = 596
( x + 58x) x [( 58x - x ) : 3 + 1 ] : 2 = 596
x = 590 = 596
x = 596 : 590
x = 596/590
Tìm x,biết:
a) x^2 - 4x -5 = 0
b) 4x^2 + 7x - 11 = 0
c) -7x^2 + 6x + 1 = 0
d) - 10x^2 +7x+3 = 0
a) x2 - 4x - 5 = 0
=> x2 - 5x + x - 5 = 0
=> x(x - 5) + (x - 5) = 0
=> (x + 1)(x - 5) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
b) 4x2 + 7x - 11 = 0
=> 4x2 + 11x - 4x - 11 = 0
=> x(4x + 11) - (4x + 11) = 0
=> (x - 1)(4x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)
c) -7x2 + 6x + 1 = 0
=> -7x2 + 7x - x + 1 = 0
=> -7x(x - 1) - (x - 1) = 0
=> (-7x - 1)(x - 1) = 0
=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)
d) -10x2 + 7x + 3 = 0
=> -10x2 + 10x - 3x + 3 = 0
=> -10x(x - 1) - 3(x - 1) = 0
=> (-10x - 3)(x - 1) = 0
=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)
\(a,x^2-4x-5=0\)
\(\Rightarrow x^2-5x+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
\(b,4x^2+7x-11=0\)
\(\Rightarrow4x^2-4x+11x-11=0\)
\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)
\(c,-7x^2+6x+1=0\)
\(\Rightarrow-7x^2+7x-x+1=0\)
\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
\(d,-10x^2+7x+3=0\)
\(\Rightarrow-10x^2+10x-3x+3=0\)
\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)
a)4x2+4x+1-x2-10x-25=0
b)(x2+x+7)(x2+x-7)=(x2+x)2-7x
a)4x2+4x+1-x2-10x-25=0
`<=>(2x+1)^2-(x+5)^2=0`
`<=>(2x+1-x-5)(2x+1+x+5)=0`
`<=>(x-4)(3x+6)=0`
`<=>(x-4)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
b)(x^2+x+7)(x^2+x-7)=(x2+x)2-7x
`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`
`<=>-7^2=-7x`
`<=>-49=-7x`
`<=>x=7`
Vậy x=7
x+2x+3x+3x+4x+5x+6x+7x+8x+9x+10x=-165.tim x
X*(1+2+3+4+5+6+7+8+9+10)=-165
X*55=-165
X=-165/55=-3
111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
(1x + 9x) + (2x + 8x) + (3x + 7x) + (4x + 6x) + 5x + 10x = -165
10x^6 + 5x = -165
= 65x (-165)
= -100
Tính giá trị của biểu thức
a) A=x3-30x2-31+1 tại x=31
b) B=x5-15x4+16x3-29x2+13x tại x=14
c) C=x14-10x13+10x12-10x11+...+10x2-10x+10 tại x=9
d) D=7x(x-5)+3(x-2) tại x=0
e) E=5-4x(x-2)+4x2 tại x=4
f) F=4x2-2x+3x(x-5) tại x=-1
g) G=(5x-7)(7x+3)-(7x+2)(5x-4) tại x=-3
h) H=(-5x+4)(3x-2)+(-2x+3)(x-2) tại x=-2
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
x.1x.2x.3x.4x.5x.6x.7x.8x.9x.10x.11x=5500
x=?
-10x^2 + 11x + 6
- 10x^2 + 4x + 6
- 10x^2 -7x + 6
- 10 x^2 -17x + 6
-10x^2 -28 - 6
-10x^2 + 7x + 12
Phân tích thành nhân tử ( bằng kĩ thuật tách hạng tử)
Tìm x nếu
(x + 2) + (4x + 4) + (7x + 6) + (10x + 8) + ... +(25x + 18) + (28x + 20) = 1560.
1. TÌm x:
a)4x^2-2x+3-4x.(x-5)=7x-3
b)-3x.(x-5)+5.(x-1)+3x^2=4x
c)7x.(x-2)-5.(x-1)=21x^2-14x^2+3
d)3.(5x-1)-x.(x-2)+x^2-13x=7
e) 1/5x.(10x-15)-2x.(x-5)=12
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~