X=5cosx-2*cos(x+pi)+tan(3/2pi-x)+7*sin(pi/2-x)

=5cosx+7cosx+2cosx-cot(pi/2-x)

=14cosx-tanx

A=x³+1+2x+2-x³-2x=3

B=5x²+36x+7-5x²+5x=41x+7

Có / x - 2 / = x - 2 hoặc = - ( x - 2 )

TH1 : Nếu / x - 2 / = x - 2 thay vao B ta có :

B = ( x - 2 ) + ( 7 - x )

   = x - 2 + 7 - x

   = x + ( 7 - 2 ) - x

   = x + 5  - x

  =  ( x - x ) + 5

  =   5

TH2 : Nếu / x - 2 / = - ( x-2 ) thay vào B ta có :

B = - (x - 2 ) + ( 7 - x )

   =  - x + 2 + 7 - x

  =  - x + 9 - x

  = ( - x - x ) + 9

 =  -(2.x )+ 9

Sai thì tớ xin lổi nhé !

= 9

Tớ không may gõ nhầm , bỏ = 9 đi nhé ! hiiiiiiiii.....

sorry mình chỉ mới có lớp 5 thôi

\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)

\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)

\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)

\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)

\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)

\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)

Làm tiếp bài của Song Ngư (๖ۣۜO๖ۣۜX๖ۣۜA)

\(D=\frac{\frac{-188}{11}x^{17}y^3}{6z^4a}\)

= 4x²-4x+1+x²+3x+9-5(x²-49)

=5x²-x+10-5x²+245

=-x+255

\(B=2.\left(x+1\right)\left(x-1\right)+\left(x+1\right)^2+\left(x-1\right)^2\)

\(B=2.\left(x^2-1\right)+\left(x^2+2x+1\right)+\left(x^2-2x+1\right)\)

\(B=2x^2-2+x^2+2x+1+x^2-2x+1\)

\(B=4x^2\)

Thay x=-3 vào B:

\(B=4.\left(-3\right)^2=4.9=36\)

Vậy B=36

7:

a: =>0,5x-5=2 hoặc 0,5x-5=-2

=>0,5x=3 hoặc 0,5x=7

=>x=6 hoặc x=14

b: |5x-2|=-3

mà |5x-2|>=0

nên ptvn

c: =>1/4x+3=0

=>1/4x=-3

=>x=-12

\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)

\(B=\dfrac{7a-7b+8a+8b-16b}{\left(a+b\right)\left(a-b\right)}=\dfrac{15a-15b}{\left(a-b\right)\left(a+b\right)}\\ B=\dfrac{15\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}=\dfrac{15}{a+b}\)