Bài 3:

a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)

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

Dễ thấy \(VT\ge0\forall x;y\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3y+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\frac{-4}{3}\end{matrix}\right.\)

Vậy...

b) \(3x^2+y^2+10x-2xy+26=0\)

\(\Leftrightarrow x^2-2xy+y^2+2x^2+10x+26=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x^2+5x+\frac{25}{4}\right)+\frac{27}{2}=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x+\frac{5}{2}\right)^2=\frac{-27}{2}\)

Dễ thấy \(VT\ge0\forall x;y\) mặt khác \(VP< 0\)

Do đó pt vô nghiệm

Bài 2:

\(A=263^2+74\cdot263+37^2\)

\(A=263^2+2\cdot263\cdot37+37^2\)

\(A=\left(263+37\right)^2\)

\(A=300^2\)

\(A=90000\)

b) tương tự

\(C=-1^2+2^2-3^2+...-99^2+100^2\)

\(C=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)

\(C=\left(2-1\right)\left(1+2\right)+\left(4-3\right)\left(3+4\right)+...+\left(100-99\right)\left(99+100\right)\)

\(C=1+2+3+4+...+99+100\)

\(C=\frac{\left(100+1\right)\cdot100}{2}=5050\)

\(D=\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)

\(2D=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)

\(2D=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)

\(2D=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)

\(2D=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)

\(2D=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(2D=\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(2D=3^{64}-1\)

\(D=\frac{3^{64}-1}{2}\)

a: \(A=\dfrac{\left(258-242\right)\left(258+242\right)}{\left(254-246\right)\left(254+246\right)}=\dfrac{16}{8}=2\)

b: \(=\left(263+37\right)^2=300^2=90000\)

c: \(=\left(136-46\right)^2=90^2=8100\)

d: \(=50^2-49^2+48^2-47^2+...+2^2-1^2\)

=50+49+...+2+1

=51x50:2=1275

b) \(263^2+74.263+37^2\)

\(=\left(263+37\right)^2\)

\(=300^2\)

\(=90000\)

c) \(136^2-92.136+46^2\)

\(=\left(136-46\right)^2\)

\(=90^2\)

\(=8100\)

cảm ơn lê tài bảo châu nhé

Trả lời:

Bài 4:

b, B =  ( x + 1 ) ( x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x - 1 ) 

= x⁸ - x⁷ + x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x - 1 

= x⁸ - 1

Thay x = 2 vào biểu thức B, ta có:

2⁸ - 1 = 255

c, C = ( x + 1 ) ( x⁶ - x⁵ + x⁴ - x³ + x² - x + 1 ) 

= x⁷ - x⁶ + x⁵ - x⁴ + x³ - x² + x + x⁶ - x⁵ + x⁴ - x³ + x² - x + 1

= x⁷ + 1

Thay x = 2 vào biểu thức C, ta có:

2⁷ + 1 = 129

d, D = 2x ( 10x² - 5x - 2 ) - 5x ( 4x² - 2x - 1 ) 

= 20x³ - 10x² - 4x - 20x³ + 10x² + 5x

= x

Thay x = - 5 vào biểu thức D, ta có:

D = - 5

Bài 5: 

a, A = ( x³ - x²y + xy² - y³ ) ( x + y )

= x⁴ + x³y - x³y - x²y² + x²y² + xy³ - xy³ - y⁴

= x⁴ - y⁴

Thay x = 2; y = - 1/2 vào biểu thức A, ta có:

A = 2⁴ - ( - 1/2 )⁴ = 16 - 1/16 = 255/16

b, B = ( a - b ) ( a⁴ + a³b + a²b² + ab³ + b⁴ ) 

= a⁵ + a⁴b + a³b² + a²b³ + ab⁴ - ab⁴ - a³b² - a²b³ - ab⁴ - b⁵ 

= a⁵ + a⁴b - ab⁴ - b⁵

Thay a = 3; b = - 2 vào biểu thức B, ta có:

B = 3⁵ + 3⁴.( - 2 ) - 3.( - 2 )⁴ - ( - 2 )⁵ = 243 - 162 - 48 + 32 = 65

c, ( x² - 2xy + 2y² ) ( x² + y² ) + 2x³y - 3x²y² + 2xy³ 

= x⁴ + x²y² - 2x³y - 2xy³ + 2x²y² + 2y⁴ + 2x³y - 3x²y² + 2xy³

= x⁴ + 2y⁴

Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:

( - 1/2 )⁴ + 2.( - 1/2 )⁴ = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16

\(\frac{63^2-47^2}{215^2-105^2}=\)  \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

                           \(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)

các câu kia làm tương tự nha

Thanks bạn nhiều nhiều nha!

nhưng câu d với câu e mk khô hiểu lắm!

A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

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

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

B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

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

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

\(=\left(x+3-1\right)\left(x+3+1\right)\)

\(=\left(x+2\right)\left(x+4\right)\)

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

x=0

x=2