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2 tháng 8 2020

Bài 4 : Tính nhanh :
a, 15. 64 + 25. 100 + 36. 15 + 60. 100

= (15 . 64 + 36. 15) + (25. 100 + 60. 100)

= 15.(64 + 36) + 100.(25 + 60)

= 15. 100 + 100. 85

= 100.(15 + 85)

= 100. 100

= 10000
b, 472 + 482 - 25 + 94. 48

= 472 + 2.47. 48 + 482 - 25

= (47 + 48)2 - 52

= (47 + 48 - 5)(47 + 48 + 5)

= (48 + 22)(48 + 52)

= 90. 100

= 9000
c, 93 - 92. ( -1) - 9. 11 + ( -1). 11

= 93 + 92 + 11(- 9 - 1)

= 92.(9 + 1) + 11. (-10)

= 81. 10 - 110

= 810 - 110

= 700
d,2016. 2018 - 20172

= (2017 - 1)(2017 + 1) - 20172

= 20172 - 1 - 20172

= -1

#Học tốt!

30 tháng 7 2020

a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)

c) \(73^2-13^2-10^2+20.13\)

\(=73^2-\left(13^2+10^2-20.13\right)\)

\(=73^2-\left(13^2-2.13.10+10^2\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2\)

\(=\left(73-3\right)\left(73+3\right)\)

\(=70.76\)

\(=5320\)

d)Viết đề = công thức trực quan hộ mình

3 tháng 8 2020

Bài 1 : Tìm x,biết :
a, x2(x + 5) - 9x = 45

⇔ x2(x + 5) - 9x - 45 = 0

⇔ x2(x + 5) - 9(x + 5) = 0

⇔ (x + 5)(x2 - 9) = 0

⇔ (x + 5)(x - 3)(x + 3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)

Vậy x ={-5; 3; -3}
b, 9(5 - x) + x2 - 10x = -25

⇔ 45 - 9x + x2 - 10x + 25 = 0

⇔ x2 - 19x + 70 = 0

⇔ x2 - 14x - 5x + 70 = 0

⇔ (x2 - 5x) - (14x - 70) = 0

⇔ x(x - 5) - 14(x - 5) = 0

⇔ (x - 5)(x - 14) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)

Vậy x ={5; 14}

3 tháng 8 2020

a, x2( x+5 ) - 9x = 45

x3 + 5x2 - 9x - 45 = 0

x2( x+5 ) - 9( x+5) = 0

(x2 - 9)(x + 5) = 0

(x + 3)(x - 3)(x + 5) = 0

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)

b, 9( 5-x ) + x2 -10x = -25

45 - 9x + x2 - 10x + 25 = 0

x2 - 19x + 70 = 0

x2 - 14x - 5x + 70 = 0

x( x-14 ) - 5( x-14) = 0

(x - 5)(x - 14) = 0

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)

30 tháng 7 2020

cái cuối hằng đẳng thức là xong mà bạn

30 tháng 7 2020

a) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

b) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left[3\left(x+y-1\right)\right]^2-\left[2\left(2x+3y+1\right)\right]^2\)

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

\(=\left(3x+3y-3+4x+6y+2\right)\left(3x+3y-3-4x-6y-2\right)\)

\(=\left(7x+9y-1\right)\left(-x-3y-5\right)\)

c) \(-4x^2+12xy-9y^2+25\)

\(=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+5^2\)

\(=-\left[\left(2x\right)^2-2.2x.3y+\left(3y\right)^2-5^2\right]\)

\(=-\left[\left(2x-3y\right)^2-5^2\right]\)

\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)

d) \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-4m\left(m-n\right)-n^2\)

\(=\left(x-y\right)^2-4m\left(m-n\right)-n^2\)

\(=\left(x-y-n\right)\left(x-y+n\right)-4m\left(m-n\right)\)

28 tháng 6 2018

giúp mk vs

28 tháng 6 2018

bucminh

5 tháng 8 2020

Bài 9 : Tìm x, biết :

a, (x - 2)(x - 3) + (x - 2) - 1 = 0

\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy x ={1; 3}

b, (x + 2)2 - 2x(2x + 3) = (x + 1)2

\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x

\(\Leftrightarrow6x^3+x^2-2x=0\)

\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)

\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)

\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)

\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)