a: ΔABC cân tại A

mà AD là đường cao

nên D là trung điểm của BC

ΔADB vuông tại D

=>\(DA^2+DB^2=AB^2\)

=>\(DB=\sqrt{5^2-4^2}=3\left(cm\right)\)

b: Xét ΔHDB vuông tại D và ΔHEA vuông tại E có

\(\widehat{DHB}=\widehat{EHA}\)(hai góc đối đỉnh)

Do đó: ΔHDB~ΔHEA

=>\(\dfrac{HD}{HE}=\dfrac{HB}{HA}\)

=>\(HD\cdot HA=HB\cdot HE\)

Giúp mình làm vs

\(\dfrac{x^2+4+6}{5-6x}< 0\)

Nhận xét:

\(x^2\ge0,\forall x\)

\(\Rightarrow x^2+4+6\ge10,\forall x\)

Do đó \(\dfrac{x^2+4+6}{5-6x}< 0\) khi và chỉ khi:

\(5-6x< 0\)

\(\Leftrightarrow6x>5\)

\(\Leftrightarrow x>\dfrac{5}{6}\)

Vậy \(x>\dfrac{5}{6}\)

Sửa đề:

   Do (x + 2)² ≥ 0

(x + 2)² + 2 > 0

5 - 6x < 0

-6x < -5

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

Nhận xét:

\(\sqrt{3}>\sqrt{1}=1\)

\(\Rightarrow1-\sqrt{3}< 0\)

\(\Rightarrow\sqrt{x}< 0\) (vô lí)

Vậy không tìm được giá trị x thoả mãn đề bài

\(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\\ =\left[\left(a+1\right)\left(a+4\right)\right]\left[\left(a+2\right)\left(a+3\right)\right]+1\\ =\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\\ =\left(a^2+5a+4\right)\left(a^2+5a+4\right)+2\left(a^2+5a+4\right)+1\\ \left(a^2+5a+4\right)^2+2\left(a^2+5a+4\right)+1\\ =\left(a^2+5a+5\right)^2\)

a) $-(2x-4)(x+2)+(x+2)^2+(x-2)^2-4x^2-1-4x=-3$

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

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

$\Leftrightarrow (2x)^2-4x^2-4x=-2$

$\Leftrightarrow -4x=-2$

$\Leftrightarrow x=\frac12$

b) $(4x-1)^2-16(x+1)(x+3)=25$

$\Leftrightarrow (4x)^2-2.4x.1+1^2-16(x^2+4x+3)=25$

$\Leftrightarrow 16x^2-8x+1-16x^2-64x-48=25$

$\Leftrightarrow -72x-47=25$

$\Leftrightarrow -72x=72$

$\Leftrightarrow x=-1$

c) $(3x-7)^2=9(3x-7)(x+5)+694$

$\Leftrightarrow (3x)^2-2.3x.7+7^2=9(3x^2+8x-35)+694$

$\Leftrightarrow 9x^2-42x+49=27x^2+72x-315+694$

$\Leftrightarrow 18x^2+114x+330=0$

$\Leftrightarrow 9x^2+57x+165=0$

$\Leftrightarrow 9\left(x+\frac{19}{6}\right)^2+\frac{299}{4}=0$ (vô lí)

=> Pt vô nghiệm

d) $(2x-1)^2+(x+3)^2=5(x+7)(x-7)-3x$

$\Leftrightarrow 4x^2-4x+1+x^2+6x+9=5(x^2-49)-3x$

$\Leftrightarrow 5x^2+2x+10=5x^2-3x-245$

$\Leftrightarrow 5x=-255$

$\Leftrightarrow x=-51$

#$\mathtt{Toru}$

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

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

=>\(-2x^2+8-2x^2-4x+7+3=0\)

=>\(-4x^2-4x+18=0\)

=>\(x=\dfrac{-1\pm\sqrt{19}}{2}\)

b: \(\left(4x-1\right)^2-16\left(x+1\right)\left(x+3\right)=25\)

=>\(16x^2-8x+1-16\left(x^2+4x+3\right)-25=0\)

=>\(16x^2-8x-24-16x^2-64x-48=0\)

=>-72x-72=0

=>x=-1

c: \(\left(3x-7\right)^2=9\left(3x-7\right)\left(x+5\right)+694\)

=>\(9\left(3x^2+15x-7x-35\right)+694=9x^2-42x+49\)

=>\(27x^2+72x-315+694-9x^2+42x-49=0\)

=>\(18x^2+114x+330=0\)

=>\(x\in\varnothing\)

d: \(\left(2x-1\right)^2+\left(x+3\right)^2=5\left(x+7\right)\left(x-7\right)-3x\)

=>\(4x^2-4x+1+x^2+6x+9=5\left(x^2-49\right)-3x\)

=>\(5x^2+2x+10-5x^2+245+3x=0\)

=>5x+255=0

=>x+51=0

=>x=-51

Bài 7

1) 

\(A=8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)

2)  

\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\\ =-\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\\ =-\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\\ =-\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =-\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =-\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =-\left(3^{32}-1\right)\\ =1-3^{32}\)