\(P=\left[\frac{a+3\sqrt{a}+2}{a+\sqrt{a}-2}-\frac{a+\sqrt{a}}{a-1}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)

\(P=\left(\frac{a+2\sqrt{a}+\sqrt{a}+2}{a+2\sqrt{a}-\sqrt{a}-2}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}}{\sqrt{a}-1}\right]:\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(P=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}}{\sqrt{a}-1}\right)\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(P=\frac{\sqrt{a}}{\sqrt{a}-1}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}=\frac{\sqrt{a}+1}{2}\)

Trả lời:

a, x² + 2y² - 2xy - 2x + 2 

= x² + y² + y² - 2xy - 2x + 1 + 1 + 2y - 2y

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

= [ ( x - y )² - 2 ( x - y ) + 1 ] + ( y - 1 )²

= ( x - y - 1 )² - ( y - 1 )² 

lam b, c di

lớp 7a trồng đc số cây là

       1020:[8+9]x9=540[cây]

.......7b........................là

        1020-540=480 [cây]

             Đs:chúc  bn   học tốt nha

Đề sai, nãy giờ cứ nghĩ hoài luôn

a, \(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

b, bạn kiểm tra lại đề 

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

\(=-3\left(x^2-2.\frac{1}{6}x+\frac{1}{36}+\frac{35}{36}\right)=-3\left(x-\frac{1}{6}\right)^2-\frac{35}{12}\le-\frac{35}{12}< 0\forall x\)

45²+40²-15²+80.45

=(45²+80,45+40²)-15²

=(45²+2.40.45+40²)-15²

=(45+40)²-15²

=(45+40-15)(45+40+15)

=70.100

=7000

#H

(B có thể dùng máy tính thử lại, kết quả chính xác là 7000)

45^2+40^2-15^2+80*45

=(45+40-15)^2+80.45

=4900+80.45

=4900+3600

=8500

Ta có: a + b + c = 0

<=> a² + b² + c² + 2(ab + bc + ac) = 0

<=> a² + b² + c² = -2(ab + bc + ac)

<=> a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + a²c² = 4[a²b² + b²c² + a²c² + 2abc(a + b + c)] (vì a + b + c= 0)

<=> a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + a²c²) = 4(a²b² + b²c² + a²c²)

<=> a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + a²c²) (đpcm)

b) Từ a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + a²c²)

<=> (a⁴ + b⁴ + c⁴)/2 = a²b² + b²c² + a²c² + 2abc(a + b + c) (vì a + b + c) = 0

<=> (a⁴ + b⁴ + c⁴)/2 = (ab + bc + ac)²

<=> a⁴ + b⁴ + c⁴ = 2(ab + bc + ac)² (đpcm)

c) Từ a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + a²c²)

<=> 2(a⁴ + b⁴ + c⁴) = a⁴+ b⁴ + c⁴ + 2(a²b² + b²c² + a²c²)

<=> 2(a⁴ + b⁴ + c⁴) = (a² + b² + c²)²

<=> a⁴ + b⁴ + c⁴ = (a² + b² + c²)²/2 (đpcm) 

Ta có a⁴ + ab³ - a³b - b⁴ 

 = (a⁴  - a³b) + (ab³ - b⁴) 

= a³(a - b)  + b³(a - b) 

= (a - b)(a³ + b³)

= (a - b)(a + b)(a² - ab  + b²)

Trả lời:

a⁴ + ab³ - a³b - b⁴

= ( a⁴ + ab³ ) - ( a³b + b⁴ )

= a ( a³ + b³ ) -  b ( a³ + b³ )

= ( a³ + b³ ) ( a - b )

= ( a + b ) ( a² + ab + b² ) ( a - b )