bài tập tết nâng cao phải ko

mk cũng có nhưng chưa làm dc

5x.(-x)²+1=6

5x.x+1=6

5x²+1=6

5x²=6-1

5x²=5

x²=5:5

x²=1

x²=1²

=>x=1

(15-x)+(x-12)=7-(-8+x)

15-x+x-12=7+8-x

3-x+x=15-x

3=15-x

x=15-3

x=12

4x³=4x

Để 4x³=4x=>x³=x

=>x=1

Câu cuối cùng mình ko bik.

hoc tốt

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ( x + 4 )² - ( x + 1 )( x - 1 ) = 16

<=> x² + 8x + 16 - ( x² - 1 ) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = -1/8

b) ( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 49 ) = 0

<=> 5x² + 2x + 10 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> 2x = -255/2

c) ( x + 2 )( x² - 2x + 4 ) - x( x² + 2 ) = 15

<=> x³ + 2³ - x³ - 2x = 15

<=> 8 - 2x = 15

<=> 2x = -7

<=> x = -7/2

d) ( x + 3 )³ - x( 3x + 1 )² + ( 2x + 1 )( 4x² - 2x + 1 ) = 28

<=> x³ + 9x² + 27x + 27 - x( 9x² + 6x + 1 ) + [ ( 2x )³ + 1³ ] = 28

<=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

<=> 3x² + 26x + 28 = 28

<=> 3x² + 26x = 0

<=> x( 3x + 26 ) = 0

<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)

\(5x\left(x-3\right)\left(x-1\right)-4x\left(x^2-2x\right)\)

\(5x^3-5x^2-15x^2+15x-4x^3+8x^2\)

\(x^3-12x^2+15x\)

\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)

\(-4x^3+16x^2-12x^2+48x-3x^3+3x^2-3x\)

\(-7x^3+7x^2+45x\)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)