#)Giải :

a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)

c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

 \(\left(5x+1\right)^2=\frac{6^2}{7^2}\)

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(\Leftrightarrow5x+1=\frac{6}{7}\)

\(5x=\frac{6}{7}-1\)

\(5x=\frac{6}{7}-\frac{7}{7}\)

\(5x=-\frac{1}{7}\)

\(x=-\frac{1}{7}\div5\)

\(x=-\frac{1}{7}\times\frac{1}{5}\)

\(x=-\frac{1}{35}\)

Vậy \(x=-\frac{1}{35}\)

a) Ta có: \(P\left(x\right)=5x^2-7+6x-8x^3-x^4\)

\(=-x^4-8x^3+5x^2+6x-7\)

Ta có: \(Q\left(x\right)=x^4+5+8x^3-5x^2\)

\(=x^4+8x^3-5x^2+5\)

b) Ta có: P(x)+Q(x)

\(=-x^4-8x^3+5x^2+6x-7+x^4+8x^3-5x^2+5\)

\(=6x-2\)

c) Ta có: P(x)-Q(x)

\(=-x^4-8x^3+5x^2+6x-7-\left(x^4+8x^3-5x^2+5\right)\)

\(=-x^4-8x^3+5x^2+6x-7-x^4-8x^3+5x^2-5\)

\(=-2x^4-16x^3+10x^2+6x-12\)

câu a dễ bạn tự làm nha

b) Ta có: P(x)+Q(x)

c) Ta có: P(x)-Q(x)

a) A(x) = 0

=> 6x + 3 - (2x + 1)

=> 6x + 3 - 2x - 1 = 0

=> (6x - 2x) + (3 - 1) = 0

=> 4x + 2 = 0

=> 4x = -2

=> x = -2 : 4

=> x = -0,5

Vậy ...

b) B(x) = 0

=> (x² + 5x - 5) - (5x - 5) = 0

=> x² + 5x - 5 - 5x + 5 = 0

=> x² + 5x - 5x = 0

=> x² = 0

=> x = 0

Vậy ...

c) C(x) = x² - 8x

=> x² - 8x = 0

=> x² = 8x

=> x = 8 ( Chia mỗi bên cho x)

Vậy ...

d) D(x) = x² - 5x + 4

=> x² - x - 4x + 4 = 0

=> x.(x - 1) - 4.(x - 1) = 0

=> (x - 4).(x - 1) = 0

=> \(\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

Vậy x = 4; x = 1 là nghiệm của D(x)

P(x)+Q(x)=(5x³-+7x⁴+8x²)+(8x²-5x-3x³+x⁴)

= 5x³-7x⁴+8x²+8x²-5x-3x³+x⁴

=(5x³-3x³)+(-7x⁴+x⁴)+(8x²+8x²)-5x

=2x³-6x⁴+16x²-5x

P(x)-Q(x)=(5x³-+7x⁴+8x²)-(8x²-5x-3x³+x⁴)

= 5x³-+7x⁴+8x²-8x²+5x+3x³-x⁴

=(5x³+3x³)+(-7x^4_x⁴)+(8x^2-8x²)+5x

= 8x³-8x⁴+5x

*Cách 1: Hàng ngang:

P(x) - Q(x) = (5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\)) - (8x\(^2\) - 5x - 3x\(^3\) + x\(^4\) - \(\dfrac{2}{3}\))

= 5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\) - 8x\(^2\) + 5x + 3x\(^3\) - x\(^4\) +\(\dfrac{2}{3}\)

= (5x\(^3\) + 3x\(^3\)) + (-\(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)) + (7x\(^4\) - x\(^4\)) + (8x\(^2\) - 8x\(^2\)) + 5x

= 8x\(^3\) + \(\dfrac{1}{3}\) + 6x\(^4\) + 5x

Vậy P(x) - Q(x) = 8x\(^3\) + \(\dfrac{1}{3}\) + 6x\(^4\) + 5x

*Cách 2: Hàng dọc:

P(x) = 7x\(^4\) + 5x\(^3\) + 8x\(^2\) + 0x - \(\dfrac{1}{3}\)

-

Q(x) = x\(^4\) - 3x\(^3\) + 8x\(^2\) - 5x - \(\dfrac{2}{3}\)

Hay: P(x) = 7x\(^4\) + 5x\(^3\) + 8x\(^2\) + 0x - \(\dfrac{1}{3}\)

+

[-Q(x)] = -x\(^4\) + 3x\(^3\) - 8x\(^2\) + 5x + \(\dfrac{2}{3}\)

___________________________________________

P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)

Vậy P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)

Ta có: A(x) = -4x⁵ - x³ + 4x² + 5x + 9 + 4x⁵ - 6x² - 2

A(x) = (-4x⁵ + 4x⁵) - x³ + (4x² - 6x²) + 5x + (9 - 2)

A(x) = -x³ - 2x² + 5x + 7

B(x) = -3x⁴ - 2x³ + 10x² - 8x + 5x³ - 7 - 2x³ + 8x

B(x) = -3x⁴ - (2x³ - 5x³ + 2x³) + 10x² - (8x - 8x) - 7

B(x) = -3x⁴ + x³ + 10x² - 7

A(x) + B(x) = (-x³ - 2x² + 5x + 7) + (-3x⁴ + x³ + 10x² - 7)

  = -x³ - 2x² + 5x + 7 - 3x⁴ + x³ + 10x² - 7

 = (-x³ + x³) - (2x² - 10x²) + 5x + (7 - 7)

 = 8x² + 5x

A(x) - B(x) = (-x^3 - 2x^2 + 5x + 7) - (-3x^4 + x^3 + 10x^2 - 7)

= -x^3 - 2x^2 + 5x + 7 + 3x^4 - x^3 - 10x^2 + 7

= (-x^3 - x^3) - (2x^2 + 10x^2) + 5x + (7 + 7)

= -2x^3 - 12x^2 + 5x + 14