Bài 1:

(a + b)² - 4ab

= a² + 2ab + b² - 4ab

= a² - 2ab +b² 

= (a-b)² (đpcm)

Bài 2:

    \(x^3\) - 9\(x^2\) + 27\(x\) - 27

=  (\(x\) - 3)³  (1)

Thay \(x\) = 5 vào (1) ta có: (5-3)³ = 8

\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\) 

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

     \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -1; 10}

\(a,15x-5xy\\ =5x\left(3-y\right)\\ b,\left(x^2+1\right)^2-4x^2\\ =\left(x^2-x+1\right)\left(x^2+x+1\right)\\ c,x^2-10x-9y^2+25\\ =\left(x-5\right)^2-9y^2\\ =\left(x-9y-5\right)\left(x+9y-5\right)\)

hé lu ông zà

 Ta có \(HN\perp AC\) và \(AB\perp AC\) nên AB//HN. Do đó tứ giác ABHN là hình thang        (1)

 Mặt khác, tam giác ABC vuông tại A có trung tuyến AM nên \(AM=\dfrac{1}{2}BC=BM\), suy ra tam giác MAB cân tại M hay \(\widehat{ABH}=\widehat{NAB}\)           (2)

 Từ (1) và (2), ta suy ra tứ giác ABHN là hình thang cân. (đpcm)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\) (1)

Do \(\left(2x-3\right)^2\ge0\) và \(\left(y+5\right)^2\ge0\)

\(\Rightarrow\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

\(2x^2-2xy-4x+y^2+4=0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-4x+4=0\)

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

mà \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0,\forall x;y\\\left(x-2\right)^2\ge0,\forall x\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

4\(x^2\) + y² - 12\(x\) + 10y + 34 = 0

(4\(x^2\) - 12\(x\) + 9) + (y² + 10y + 25) = 0

(2\(x\) - 3)² + (y + 5)² = 0

(2\(x\) - 3)² ≥ 0 ∀ \(x\); (y + 5)² ≥ 0 ∀ y

(2\(x-3\))² + (y + 5)² = 0 ⇔ \(\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Kl: (\(x;y\)) = ( \(\dfrac{3}{2}\); -5)

\(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

mà \(\left\{{}\begin{matrix}\left(2x-3\right)^2\ge0,\forall x\\\left(y+5\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Ta có : \(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

Ta thấy : \(\left(2x-3\right)^2;\left(y+5\right)^2\ge0\)

Nên để (1) thoả mãn : 

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Vậy........

\(5^{x+1}-5^x=100\cdot25^{29}\)

\(\Rightarrow5^x\left(5-1\right)=100\cdot\left(5^2\right)^{29}\)

\(\Rightarrow5^x\cdot4=100\cdot5^{58}\)

\(\Rightarrow5^x=\dfrac{100\cdot5^{58}}{4}\)

\(\Rightarrow5^x=25\cdot5^{58}\)

\(\Rightarrow5^x=5^{60}\)

\(\Rightarrow x=60\)

\(5^{x+1}-5x=100.25^{29}\)

\(5.5^x-5^x=4.25.25^{29}\)

\(5^x.\left(5-1\right)4.25^{30}\)

\(4.5^x-4.\left(5^2\right)^{30}\)

\(5x=5^{60}\)

\(x=60\)

tham khảo nhé, phần nào thừa thì bạn có thể ko vt

Góc A = 3. góc D 

góc A + góc D = 1800

Giải bài toán tổng tỉ trên ta được :

góc A = 180:(1+3).3=1350 

góc B - góc C = 30

góc B + góc C = 1800

Giải bài toán tổng hiệu trên ta được : 

Góc B = ( 180+30 ) :2 = 1050

Tổng : góc A + góc B = 1350+1050= 2040