Tính :
4 x 4 = ?
2 x 2 = ?
3 x 2 = ?
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Sửa đề: Các dấu bằng ở yêu cầu là dấu cộng.
1. Có: \(x+y=3\)
\(\Leftrightarrow\left(x+y\right)^2=3^2\)
\(\Leftrightarrow x^2+2xy+y^2=9\)
\(\Leftrightarrow x^2+y^2=9-2\cdot1=7\) (do \(xy=1\))
\(------\)
Lại có: \(x+y=3\)
\(\Leftrightarrow\left(x+y\right)^3=3^3\)
\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=27\)
\(\Leftrightarrow x^3+y^3+3\cdot1\cdot3=27\) (do x + y = 3; xy = 1)
\(\Leftrightarrow x^3+y^3=18\)
Ta có: \(x^2+y^2=7\)
\(\Leftrightarrow\left(x^2+y^2\right)^2=7^2\)
\(\Leftrightarrow x^4+y^4+2\cdot\left(xy\right)^2=49\)
\(\Leftrightarrow x^4+y^4=49-2\cdot1=47\) (do xy = 1)
4 x 2 x 5 = ?
Cách 1: 4 x 2 x 5 = (4 x 2) x 5 = 8 x 5 = 40
Cách 2: 4 x 2 x 5 = 4 x (2 x 5) = 4 x 10 = 40
7 x 2 x 3 = ?
Cách 1: 7 x 2 x 3 = (7 x 2) x 3 = 14 x 3 = 42
Cách 2: 7 x 2 x 3 = 7 x (2 x 3) = 7 x 6 = 42
6 x 3 x 3 = ?
Cách 1: 6 x 3 x 3 = (6 x 3) x 3 = 18 x 3 = 54
Cách 2: 6 x 3 x 3 = 6 x (3 x 3) = 6 x 9 = 54
6 x 2 x 4 = ?
Cách 1: 6 x 2 x 4 = (6 x 2) x 4 = 12 x 4 = 48
Cách 2: 6 x 2 x 4 = 6 x (2 x 4) = 6 x 8 = 48
a) 2 x 2 = 4 3 x 3 = 9
2 x 4 = 8 3 x 5 = 15
2 x 6 = 12 3 x 7 = 21
2 x 8 = 16 3 x 9 = 27
4 x 4 = 16 5 x 5 = 25
4 x 2 = 8 5 x 7 = 35
4 x 6 = 24 5 x 9 = 45
4 x 8 = 32 5 x 3 = 15
b) 200 x 4 = 800 300 x 2 = 600
200 x 2 = 400 300 x 3 = 900
400 x 2 = 800 500 x 1 = 500
100 x 4 = 400 100 x 3 = 300
Học sinh nhẩm và ghi kết quả như sau:
3 x 4 = 12 2 x 6 = 12 4 x 3 = 12 5 x 6 = 30
3 x 7 = 21 2 x 8 = 16 4 x 7 = 28 5 x 4 = 20
3 x 5 = 15 2 x 4 = 8 4 x 9 = 36 5 x 7 = 35
3 x 8 = 24 2 x 9 = 18 4 x 4 = 16 5 x 9 = 45
3 x4 = 12
3 x7 = 21
3 x5 = 15
3 x8 =24
2 x 6= 12
2 x8 = 16
2 x4 =8
2 x9 =18
4 x3 =12
4 x7 =28
4 x9 =36
4 x4 =16
5 x6 =30
5 x4 =20
5 x7 =35
5 x9 =45
5,5 + \(\dfrac{3}{4}\) - 5 + \(\dfrac{1}{4}\)
(5,5 - 5) + (\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\))
= 0,5 + 1
= 1,5
\(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) : \(\dfrac{3}{2}\)
= \(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) x \(\dfrac{2}{3}\)
= (\(\dfrac{5}{2}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= (\(\dfrac{10}{4}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{4}\) x \(\dfrac{2}{3}\)
= \(\dfrac{11}{6}\)
Tính \(x\):
435 - [\(x\) + 16] = 425 : 17
435 - [\(x\) + 16] = 25
[\(x\) + 16] = 435 - 25
\(x\) + 16 = 410
\(x\) = 410 - 16
\(x\) = 394
a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)
=(x-3)^2/(x-3)
=x-3
b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)
\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)
\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)
=-7(x+2)/(x-2)(x+2)
=-7/(x-2)
3 - [ \(\dfrac{2}{4}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) \(\times\) \(\dfrac{3}{2}\) + \(\dfrac{24}{8}\) \(\times\)\(\dfrac{2}{4}\) + \(\dfrac{2}{4}\)]
= 3 - \(\dfrac{2}{4}\) \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) + 3 + 1)
= 3 - \(\dfrac{2}{4}\) \(\times\) ( 2 + 3 + 1)
= 3 - \(\dfrac{1}{2}\) \(\times\) 6
= 3 - 3
= 0
Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
a: =16-2+91=14+91=105
b: =9*5+8*10-27=45+53=98
c: =32+65-3*8=8+65=73
d; \(=5^3-10^2=125-100=25\)
e: \(=4^2-3^2+1=8\)
f: =9*16-16*8-8+16*4
=16(9-8+4)-8
=16*5-8
=72
a) \(2^4-50:25+13\cdot7\)
\(=2^4-2+91\)
\(=16-2+91\)
\(=14+91\)
\(=105\)
b) \(3^2\cdot5+2^3\cdot10-3^4:3\)
\(=9\cdot5+8\cdot10-3^3\)
\(=45+80-27\)
\(=98\)
c) \(2^5+5\cdot13-3\cdot2^3\)
\(=32+65-3\cdot8\)
\(=32+65-24\)
\(=73\)
d) \(5^{13}:5^{10}-5^2\cdot2^2\)
\(=5^{13-10}-\left(5\cdot2\right)^2\)
\(=5^3-10^2\)
\(=125-100\)
\(=25\)
e) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^{5-3}-3^{9-7}+1\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=8\)
f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)
\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)
\(=4^2\cdot\left(3^2-2^3+2^2\right)-2^3\)
\(=4^2\cdot\left(9-8+4\right)-8\)
\(=16\cdot5-8\)
\(=72\)
Mình thu gọn 2 đa thức trước r mới cộng nhé
\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)
\(P\left(x\right)=\left(3x^2-3x^2\right)+\left(7-4\right)+2x^4-5x+2x^3\)
\(P\left(x\right)=2x^4+2x^3-5x+3\)
\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)
\(Q\left(x\right)=\left(-3x^3+x^3\right)+2x^2+\left(-x^4+5x^4\right)+\left(x+4x\right)-2\)
\(Q\left(x\right)=-2x^3+4x^4+2x^2+5x-2\)
\(P\left(x\right)+Q\left(x\right)=2x^4+2x^3-5x+3-2x^3+4x^4+2x^2+5x-2\)
\(P\left(x\right)+Q\left(x\right)=\left(2x^4+4x^4\right)+\left(2x^3-2x^3\right)+\left(-5x+5x\right)+\left(3-2\right)+2x^2\)
\(P\left(x\right)+Q\left(x\right)=6x^4+1+2x^2\)
4x4=16
2x2=4
3x2=6
4 x 4 = 16
2 x 2 =4
3 x 2 = 6