Ôn tập toán 8

a) 

(x³ – 3x²+3x-1) – (x³+9) +(3x²-12) = 2

3x-22 = 2

3x =24

=> x= 8

b)

x³+6x²+12x+8-6x²-x³-x²+x² = 5

                    12x+8 = 5

                      12x = -3

                       =>x = -1/4

a) ( x - 1 )³ - ( x + 3 )(x² - 3x + 9 ) + 3( x² - 4 ) = 2

(x³-3x²\(\times\)1+3x\(\times\)1²-1³)-(x³+3³)+(3x²-3\(\times\)4)=2

x³-3x²+3x-1-x³-9+3x²-12=2

3x-40=2

3x=42

x=14

b ) ( x + 2 )³ - 6x² - x² ( x + 1 ) + x² = 5

(x³+3\(\times\)2x²+3x\(\times\)2²+2³)-6x²-(x³+x²)+x²=5

x³+6x²+12x+8-6x²-x³-x²+x²=5

12x+8=5

12x=\(-\)3

x=\(-\frac{1}{4}\)

bài dễ mà 

cho đăng kí gmail you bt r` cho lớp 6 đi giải cho dễ

đăng kí

đăng kí 

(3x + 1)² - (3x - 1).(3x + 1) = 1

<=> (3x + 1).[(3x + 1) - (3x - 1)] = 1

<=> (3x + 1).(3x + 1 - 3x + 1) = 1

<=> (3x + 1).2 = 1

<=> 3x + 1 = 1/2

<=> 3x = -1/2

<=> x = -1/6

Vậy S = {-1/6}.

36x² - 25 - x.(6x - 5) = 0

<=> (36x² - 25) - x.(6x - 5) = 0

<=> [(6x)² - 5²] - x.(6x - 5) = 0

<=> (6x - 5).(6x + 5) - x.(6x - 5) = 0

<=> (6x - 5).(6x + 5 - x) = 0

<=> (6x - 5).(5x + 5) = 0

<=> 5.(6x - 5).(x + 1) = 0

<=> 6x - 5 = 0 hoặc x + 1 = 0

<=> x = 5/6 hoặc x = -1

Vậy S = {-1; 5/6}.

a)

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

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

\(\Rightarrow6x+2=1\)

\(\Rightarrow x=-\frac{1}{6}\)

Vậy pt có nghiệm là x = - 1 / 6

b)

\(36x^2-25-x\left(6x-5\right)=0\)

\(\Rightarrow\left(36x^2-25\right)-x\left(6x-5\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(6x+5\right)-x\left(6x-5\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(6x+5-x\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(5x+5\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{6}\\x=-1\end{array}\right.\)

Vậy pt có nghiệm là x = 5 / 6 ; x = - 1

\(\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)

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

\(\left(a^2+1\right)^2-6\left(a^2+1\right)+9=\left(a^2+1-3\right)^2=\left(a^2-2\right)^2\)

\(\left(x+y\right)^2+\left(x+y\right)x+\frac{1}{4}x^2=\left(x+y+\frac{1}{2}x\right)^2=\left(\frac{3}{2}x+y\right)^2\)

\(16x^4-9x^2=x^2\left(16x^2-9\right)=x^2\left(4x-4\right)\left(4x+3\right)\)

\(a^2-b^4=\left(a-b^2\right)\left(a+b^2\right)\)

(x + 2y)² - 16

= (x + 2y)² - 4²

= (x + 2y - 4).(x + 2y + 4)

(x - 2y)² - 4.(x - 2y) + 4

= (x - 2y)² - 2.(x - 2y).2 + 2²

= (x - 2y - 2)²

(a² + 1)² - 6.(a² + 1) + 9

= (a² + 1)² - 2.(a² + 1).3 + 3²

= (a² + 1 - 3)²

= (a² - 2)²

(x + y)² + (x + y).x + 1/4.x²

= (x + y)² + 2.(x + y).1/2.x + (1/2.x)²

= (x + y + 1/2.x)²

= (3/2.x + y)²

16x⁴ - 9x²

= (4x²)² - (3x)²

= (4x² - 3x).(4x² + 3x)

a² - b⁴

= a² - (b²)²

= (a - b²).(a + b²)

a. x² + x + 1

= x² + 2.x.1/2 + 1/4 + 3/4

= (x + 1/2)² + 3/4

Mà (x + 1.2)² ≥0

=> (x + 1/2)² + 3/4 ≥3/4

Vậy GTNN của đa thức là 3/4 <=> x + 1/2 = 0 <=> x = -1/2

b. (x - 1)(x + 2)(x + 3)(x + 6)

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

= (x² + 6x - x - 6)(x² + 3x + 2x + 6)

= (x² + 5x - 6)(x² + 5x + 6)

= (x² + 5x)² - 6²

= (x² + 5x)² - 36

Mà (x² + 5x)^2 ≥0

=> (x² + 5x)² - 36 ≥-36

Vậy đa thức có GTNN là -36 <=> x² + 5x = 0 <=> x.(x + 5) = 0 <=> x = 0 hoặc x + 5 = 0 <=> x = 0 hoặc x = -5.

x³ + y³ + z³ - 3xyz

= (x³ + 3x²y + 3xy² + y³) + z³ - (3x²y + 3xy² + 3xyz)

= (x + y)³ + z³ - 3xy(x + y + z)

= (x + y + z)[(x + y)² - z(x + y) + z²] - 3xy(x + y + z)

= ( x + y + z)(x² + 2xy + y² - xz - zy + z² - 3xy)

= (x + y + z)(x² + y² + z² - xy - yz - xz)

b. (x - 1)(x - 2)(x - 3)(x - 4) - 3

= (x - 1)(x - 4)(x - 2)(x - 3) - 3

= (x² - 5x + 4)(x² - 5x + 6) - 3

Đặt t = x² - 5x + 4

=> Đa thức

<=> t.(t + 2) - 3

= t² + 2t - 3

= t² + 3t - t - 3

= t.(t + 3) - (t + 3)

= (t + 3)(t + 1) (1)

Thay t = x² - 5x + 4 vào (1):

=> (x - 1)(x - 2)(x - 3)(x - 4) - 3

= (x² - 5x + 4 + 3)(x² - 5x + 4 + 1)

= (x² - 5x + 7)(x² - 5x + 5)

a: \(AB=\sqrt{AH^2+HB^2}=3\sqrt{13}\left(cm\right)\)

\(AC=\sqrt{AH^2+HC^2}=2\sqrt{13}\left(cm\right)\)

b: XétΔABC có \(BC^2=AB^2+AC^2\)

nên ΔABC vuông tại A

\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)

\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x+3\right)\left(x-3\right)}\right)\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{6x+9}{2x+3}\cdot\dfrac{1}{x-3}\)

\(=\dfrac{x-3}{x-3}=1\)