Toán

a: B=-4x^3y^2/2xy=-2x^2y

b: \(\Leftrightarrow H-2xy=\dfrac{4x^3y^2-3x^2y^3}{-2x^2y}=-2xy+\dfrac{3}{2}y^2\)

=>H=-2xy+3/2y^2+2xy=3/2y^2

\(a,4x^3y^2:B=-2xy\\ B=\dfrac{4x^3y^2}{-2xy}=4:\left(-2\right).\left(x^3:x\right).\left(y^2:y\right)=-2x^2y\\ b,\left(4x^3y^2-3x^2y^3\right):B=-2xy+H\\ \Leftrightarrow\left(4x^3y^2-3x^2y^3\right):\left(-2x^2y\right)=-2xy+H\\ \Leftrightarrow4x^3y^2:\left(-2x^2y\right)-3x^2y^3:\left(-2x^2y\right)=-2xy+H\\ \Leftrightarrow-2xy+\dfrac{3}{2}y^2=-2xy+H\\ \Rightarrow H=-2xy+2xy+\dfrac{3}{2}y^2=\dfrac{3}{2}y^2\\ Vậy:H=\dfrac{3}{2}y^2\)

a) Ta có:

\(4x^3y^2:B=2xy\)

\(\Rightarrow B=\dfrac{4x^3y^2}{-2xy}\)

\(\Rightarrow B=-2x^2y\)

b) \(\left(4x^3y^2-3x^2y^3\right):B=-2xy+H\)

\(\Rightarrow\dfrac{4x^3y^2-3x^2y^3}{-2x^2y}=-2xy+H\)

\(\Rightarrow\dfrac{-2x^2y\left(-2xy+\dfrac{3}{2}y^2\right)}{-2x^2y}=-2xy+H\)

\(\Rightarrow-2xy-\dfrac{3}{2}y^2=-2xy+H\)

\(\Rightarrow H=-2xy-\dfrac{3}{2}y^2+2xy\)

\(\Rightarrow H=-\dfrac{3}{2}y^2\)

a: =b-c-a+c+1-a-b+c

=-2a+1

b: =a-b-c-b+c+a+c-b-a

=c-3b+a

c: =2(a-b-b+c-c+a)

=2(2a-2b)

=4a-4b

a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)

\(=b-c-a+c+1-a-b+c\)

\(=c-2a+1\)

b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)

\(=a-b-c-b+c+a+c-b-a\)

\(=a-3b+c\)

c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)

\(=2\cdot\left(a-b-b+c-c+a\right)\)

\(=2\cdot\left(2a-2b\right)\)

\(=4a-4b\)

a=5*8+4=44

y=44*3=132

x=2*132=264

x:2=y

y:3=a

a:5 = 8 (dư 4)

Vậy: a= 8.5 + 4 = 44

y= a.3 = 44.3 = 132

x= y.2= 132.2= 264

Vậy: x=264

ΔABC vuông tại A có AM là trung tuyến

nên MA=MB

mà MA=AB

nên MA=AB=MB

=>ΔMAB đều

=>góc B=60 độ

=>góc C=90-60=30 độ

sin C=sin 30=1/2

a, 114 - 10 = 104 ; 166 - 10 = 156

104 = 2³ x 13 ; 156 = 2² x 3 x 13

ƯCLN(104; 156)= 2² x 13 = 4 x 13= 52

\(a\inƯ\left(52\right)=\left\{1;2;4;13;26;52\right\}\)

Vì 114 và 116 chia a dư 10 => a> 10

Vậy a=13 hoặc a=26 hoặc a=52

406 - 15 = 391 ; 498 - 15 = 483

391= 17 x 23 ; 483= 3 x 7 x 23

=> ƯCLN(391; 483)= 23

Ta có: \(a\inƯ\left(23\right)=\left\{1;23\right\}\)

Vì: 406 và 498 chia a dư 15 => a>15 

Vậy a=23

a: =>3x-15-2x+6=51

=>x-9=51

=>x=60

b: =>432-2x+22+3x-15=200

=>x+439=200

=>x=-239

c: =>3x+17-4x+13=0

=>30-x=0

=>x=30

d: =>x-3-3x+6=-11-21=-32

=>-2x+3=-32

=>-2x=-35

=>x=35/2

\(3\cdot\left(x-5\right)-2\cdot\left(x-3\right)=51\)

\(\Rightarrow3x-15-2x+6=51\)

\(\Rightarrow x-9=51\)

\(\Rightarrow x=51+9\)

\(\Rightarrow x=60\)

b) \(432-2\cdot\left(x-11\right)+3\cdot\left(x-5\right)=200\)

\(\Rightarrow432-2x+22+3x-15=200\)

\(\Rightarrow x+439=200\)

\(\Rightarrow x=200-439\)

\(\Rightarrow x=-239\)

c) \(3x+17=4x-13\)

\(\Rightarrow4x-3x=17+13\)

\(\Rightarrow x=30\)

d) \(\left(x-3\right)+21=3\cdot\left(x-2\right)-11\)

\(\Rightarrow x-3+21=3x-6-11\)

\(\Rightarrow x+18=3x-17\)

\(\Rightarrow3x-x=18+17\)

\(\Rightarrow2x=35\)

\(\Rightarrow x=\dfrac{35}{2}\)

c: A=-6x^2-10x+25x^2-35x+10x-14-19(x^2+8x+16)

=19x^2-35x-14-19x^2-152x-304

=-187x-318

Khi x=2 thì A=-318-187*2=-692

c) \(-2x\left(3x+5\right)+\left(5x+2\right)\left(5x-7\right)-19\left(x+4\right)^2\)

\(=\left(-6x^2-10x\right)+\left(25x^2-35x+10x-14\right)-19\left(x^2+8x+16\right)\)

\(=\left(-6x^2-10x\right)+\left(25x^2-25x-14\right)-\left(19x^2+152x+304\right)\)

\(=-6x^2-10x+25x^2-25x-14-19x^2-152x-304\)

\(=\left(-6x^2+25x^2-19x^2\right)-\left(10x+25x+152x\right)-\left(14+304\right)\)

\(=-187x-318\)

Thay x=2 vào BT ta có:

\(A=-187\cdot2-318=-692\)

Vậy: ...

3

\(A=sin\left(x+\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)\\ =\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{4}+x-\dfrac{\pi}{4}\right)+sin\left(x+\dfrac{\pi}{4}-x+\dfrac{\pi}{4}\right)\right]\\ =\dfrac{1}{2}\left(sin2x+sin\dfrac{\pi}{2}\right)\\ =\dfrac{1}{2}\left(-\dfrac{1}{3}+1\right)\\ =\dfrac{1}{3}\)

3:

\(A=\dfrac{\sqrt{2}}{2}\cdot\left(sinx+cosx\right)\cdot\dfrac{\sqrt{2}}{2}\cdot\left(sinx+cosx\right)\)

\(=\dfrac{1}{2}\cdot\left(sinx+cosx\right)^2\)

\(=\dfrac{1}{2}\left(1+sin2x\right)=\dfrac{1}{2}\left(1+\dfrac{-1}{3}\right)=\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{1}{3}\)

\(\left(x-1\right)^2=25\\ \Rightarrow\left(x-1\right)^2=5^2\\ \Rightarrow x-1=5\\ \Rightarrow x=6\)

\(3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)

\(2^{x-1}=32\\ \Rightarrow2^{x-1}=2^5\\ \Rightarrow x-1=5\\ \Rightarrow x=6\)

\(3^{x-2}=81\\ \Rightarrow3^{x-2}=3^4\\ \Rightarrow x-2=4\\ \Rightarrow x=6\)

\(\left(x-1\right)^2=25\)

\(\left(x-1\right)^2=5^2\)

\(x-1=5\)

\(x=6\)

\(3^x=27\)

\(3^x=3^3\)

\(=3\)

\(2^{x-1}=32\)

\(2^{x-1}=2^5\)

\(x-1=5\)

x\(x=6\)

\(3^{x-2}=81\)

\(3^{x-2}=3^4\)

\(x-2=4\)

\(x=6\)

\(M=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{4}{x-4}\left(dk:x\ge0,x\ne4\right)\)

\(=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{4}\)

\(=\dfrac{2\sqrt{x}}{4}\)

\(=\dfrac{\sqrt{x}}{2}\)

Vậy \(M=\dfrac{\sqrt{x}}{2}\) với \(x\ge0,x\ne4\)

ĐK: \(x\ge0;x\ne4\)

Khi đó:

\(M=\left(\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right).\dfrac{x-4}{4}\\ =\left(\dfrac{\sqrt{x}+2+\sqrt{x}-2}{x-4}\right).\dfrac{x-4}{4}\\ =\dfrac{2\sqrt{x}\left(x-4\right)}{\left(x-4\right).2.2}\\ =\dfrac{\sqrt{x}}{2}\)

\(ĐK:x\ne4\\ M=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{4}{x-4}\\ =\dfrac{\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{x-4}.\dfrac{\left(x-4\right)}{4}\\ =\dfrac{2\sqrt{x}\left(x-4\right)}{4\left(x-4\right)}=\dfrac{\sqrt{x}\left(x-4\right)}{2\left(x-2\right)}\)