\(A=\frac{sin^2x+cos^2x+2sinx.cosx-1}{\frac{cosx}{sinx}-sinx.cosx}=\frac{2sinx^2x.cosx}{cosx-sin^2x.cosx}=\frac{2sin^2x.cosx}{cosx\left(1-sin^2x\right)}\)

\(=\frac{2sin^2x}{1-sin^2x}=\frac{2sin^2x}{cos^2x}=2tan^2x\)

\(N=\left(\frac{sinx+\frac{sinx}{cosx}}{cosx+1}\right)^2+1=\left(\frac{sinx.cosx+sinx}{cosx\left(cosx+1\right)}\right)^2+1\)

\(=\left(\frac{sinx\left(cosx+1\right)}{cosx\left(cosx+1\right)}\right)^2+1=tan^2x+1=\frac{1}{cos^2x}\)

a. Ta có: \(3x2xy-\frac{2}{3}x^2y-4x^2.\frac{1}{3}y=6x^2y-\frac{4}{3}x^2y=\left(6-\frac{2}{3}-\frac{4}{3}\right)x^2y=4x^2y.\)

b. Thay \(x=-2,y=\frac{1}{8}\)vào đơn thức \(4x^2y\), ta được: \(4x^2y=4\left(-2\right)^2.\frac{1}{8}=2\).

Vậy, giá trị của biểu thức \(x=-2,y=\frac{1}{8}\rightarrow=2\) 

a, A = ( -a - b + c) - ( -a - b - c)

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

= 2c

b, c = -2

=> A = 2.-2 = -4

Bạn có thể làm trình bày cho mình luôn được hông

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

   = -a - b + c + a + b + c

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

   = 2c

b) Thay a = 1; b = -1; c = -2 vào A, ta có :

A = [ -1 - ( -1 ) + ( -2 ) ] - [ -1 - ( -1 ) - ( -2 ) ] 

A = -2 + 2

A = 0

Vậy A = 0 khi a = 1; b = -1; c = -2

Câu 1: C

Câu 2: D

Câu 3: A

Câu 4: B

a, A = a + ( 42 - 70 + 18 ) - ( 42 + 18 + a )

    A = a + 42 - 70 + 18 - 42 - 18 - a

    A = ( a - a ) + ( 42 - 42 ) + ( 18 - 18 ) - 70

    A = 0 + 0 + 0 - 70 

    A = -70

b, B = a + 30 + 12 - ( -20 ) + ( -12 ) - ( 2 + a )

    B = a + 30 + 12 + 20  - 12 - 2 - a

    B = ( a - a ) + ( 30 + 20 ) + ( 12 - 12 ) - 2

    B = 50 - 2

    B = 48

Chúc bạn học tốt!  

A = a + 42 -70 +18 - 42 - 18 - a

A = a + (-70) - a

A = a - 70 - a

B = a + 30 + 12 + 20 - 12 - 2 - a

B = a + 48 - a

a) A=a+(42-70+18)-(42+18+a)

    A=a+42-70+18-42-18-a

   A=a-a+42-42+18-18-70

   A=-70

b) B= a+30+12-(-20)+(-12)-(2+a)

    B=a+30 +12+20-12-2-a

    B=a-a+12-12+30+20-2

    B=50-2

   B=48

\(F=\left(1-2\sqrt{\frac{a}{b}}+\frac{a}{b}\right):\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2=\left(1-\sqrt{\frac{a}{b}}\right)^2:\left(\sqrt{a}-\sqrt{b}\right)^2\)

                                                        \(=\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)

ĐK: \(ab\ge0;b\ne0\)

\(F=\left(1-2\sqrt{\frac{a}{b}}+\frac{a}{b}\right):\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2\)

\(=\left(\sqrt{\frac{a}{b}}-1\right)^2:\left(\sqrt{a}-\sqrt{b}\right)^2=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2:\left(\sqrt{b}-\frac{b}{\sqrt{a}}\right)^2=\left(\sqrt{a}-\sqrt{b}\right)^2:\left[\frac{\sqrt{b}}{\sqrt{a}}\left(\sqrt{a}-\sqrt{b}\right)\right]^2\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2:\left[\frac{\sqrt{b}}{\sqrt{a}}\left(\sqrt{a}-\sqrt{b}\right)\right]^2\)

 \(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{a}{b\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{a}{b}\)

  ( a + b + c)³ +  (a - b - c)³ - 6a( b+ c)²

= ( a+ b + c + a - b - c)[ (a+b+c)² + (a+b+c)(a-b-c) + (a-b-c)² ] - 6a( b+c)²

= 2a [ a² + b² + c² + 2ab+ 2bc+ 2ac + a² - ( b+ c)² + a² + b² + c² - 2ab - 2ac + 2bc] - 6a ( b+c)²

= 2a [ 3a² + 2b² + 2c² + 4bc - (b+c)² - 3(b+c)²}

= 2a ( 3a² + 2b² + 2c² - 4( b+ c)² + 4bc} 

Đặt \(b+c=x\)

Biểu thức đã cho \(=\left(a+x\right)^3+\left(a-x\right)^3-6ax^2=a^3+3a^2x-3ax^2+x^3+a^3-3a^2x+3ax^2-x^3-6ax^2=2a^3\)

ko bits

a, -ac+bc

b, a+bc

Giải :

( a - b ) - ( b + c ) + ( c - a ) - ( a - b - c )

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

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

= 0 - b - ( - c )

= b + c

Học tốt !!!!!!!!!!!