Gọi hình thang đó là \(ABCD\)có \(AB\)là đáy nhỏ, \(CD\)là đáy lớn.

Khi đó \(AB=AD=BC=1\left(cm\right),AD\perp AC\).

Hạ đường cao \(AH,BK\).

Dễ thấy \(DH=CK\).

Đặt \(DH=CK=x\left(cm\right)\).

Xét tam giác \(ADC\)vuông tại \(A\)đường cao \(AH\): 

\(AD^2=DH.DC\)

\(\Leftrightarrow1=x\left(2x+1\right)\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(CD=2x+1=2\left(cm\right)\)

\(AC=\sqrt{CD^2-AD^2}=\sqrt{2^2-1}=\sqrt{3}\left(cm\right)\)

Trả lời:

1, A = x³ - 6x² + 12x - 8 = x³ - 3.x².2 + 3.x.2² - 2³ = ( x - 2 )³ 

Thay x = 22 vào A, ta được:

A = ( 22 - 2 )³ = 20³ = 8000

2, B = x² + 4x + 4 = x² + 2.x.2 + 2² = ( x + 2 )²

Thay x = 98 vào B, ta được:

B = ( 98 + 2 )² = 100² = 10000

Ta có A = x³ - 6x² + 12x - 8 = (x - 2)³

Thay x = 22 vào A 

=> A = (22 - 2)³ = 20³ = 8000

b) Sửa đề  B = x² + 4x + 4

= (x + 2)²

Thay x = 98 vào B => B = (98 + 2)²  = 100² = 10000

<=>\(1.x^2-1=3=>x^2=4=\left(+-2\right)^2=>x=+-2\)

Trả lời:

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

\(\Leftrightarrow x^2-1=3\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy S = { 2 ; - 2 }

2, \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-2\right)\left(x+2\right)=5\)

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

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x-1=5\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(S=\left\{\frac{3}{2}\right\}\)

3, 7x + 4 ( 5 - x ) = 3

<=> 7x + 20 - 4x = 3

<=> 3x + 20 = 3

<=> 3x = - 17

<=> x = - 17/3

Vậy S = { - 17/3 }

( x - 1 )³ = x³ - 3x² + 3x - 1

( x + 1 )³ = x³ + 3x² + 3x + 1

Trả lời:

( x - 1 )³ = x³ - 3.x².1 + 3.x.1² - 1³ = x³ - 3x² + 3x - 1

( x + 1 )³ = x³ + 3.x².1 + 3.x.1² + 1³ = x³ + 3x² + 3x + 1

2\(\left(x-\frac{1}{2}\right)^2=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=x^2-x+\frac{1}{4}\)

Trả lời:

1, ( x - 2 )( x² + 4x + 4 ) - ( x³ + 16 )

= x³ + 4x² + 4x - 2x² - 8x - 8 - x³ - 16

= 2x² - 4x - 24

2, \(\left(x-\frac{1}{2}\right)^2=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=x^2-x+\frac{1}{4}\)

3, \(\left(2x+5\right)^2=\left(2x\right)^2+2.2x.5+5^2=4x^2+20x+25\)

4, \(\left(-x+\frac{1}{2}\right)^2=\left(-x\right)^2+2.\left(-x\right).\frac{1}{2}+\left(\frac{1}{2}\right)^2=x^2-x+\frac{1}{4}\)

\(x^2-2x+x-2=0\)

\(x^2-x-2=0\)

\(x^2+x-2x-2=0\)

\(x\left(x+1\right)-2\left(x+1\right)=0\)

\(\left(x+1\right)\left(x-2\right)=0\)

\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

Dap an : x = - 1 , x = 2

a(b³ - c³) + b(c³ - a³) + c(a³ - b³) 

= ab³ - ac³ + bc³ - ba³ + ca³ - cb³

= b(c³ - a³ - cb² + ab²) - ac(c² - a²) 

= b[(c - a)(a² + ac + c²) - b²(c - a)] - ac(c - a)(c + a) 

= (c - a)[a²b + abc + bc² - b³ - ac² - a²c]

= (c - a)[b(c² - b²) - ac(c - b) - a²(c -b)]

= (c - a)(c - b)[b(b + c) - ac - a²] = (c - a)(c - b)(b² + bc - ac - a²]

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

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

(a + b)³ - (a - b)³ 

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

= 2b(a² + 2ab + b² + a² - b² + a² - 2ab + b²]

= 2b(3a² + b²]

x³ - 3x² + 3x - 1 - y³

= (x - 1)³ - y³ 

 = (x - y - 1)(x² - 2x + 1 + xy - y + y²]

x^{m + 4} + x^{m + 3} - x - 1 

= x^{m + 3}(x +1) - (x + 1)

= (x^{m + 3} - 1)(x + 1) 

= (x - 1)[x^{m + 2} + x^{m + 1} + .... + 1](x + 1)

(x - 1 ) ( x + 1 ) + 1

= (x + 1 )x - ( x + 1 ) + 1

= x² + x - x - 1 + 1

= x² + ( x - x ) + ( -1 + 1 )

= x²

(x - 1 ) (x + 1 ) + 1 = 0 nhe !

\(-x^2+5x+4x-20-x^2+x-3x+3=-2x^2\)

\(-2x^2+7x-17=-2x^2\)

\(7x-17=0\)

\(x=\frac{17}{7}\)