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

- Cách 1: $A=\{17;18;19;20;21;22;23\}$

- Cách 2: $A=\{x\in \mathbb{N}^*|17< x\le 23\}$

tk

c1 :A ={ 18 ;19;20;21;22;23;24}

c2 : A = { x E N / 17<x<25}

gà mái đẻ được 15 quả trứng

0 quả à

(-15)x2-240-6+36:(-6)x2

=-30-246+(-6)x2

=-276-12=-288

\(\left(-15\right)\times2-240-6+36:\left(-6\right)\times2\)

\(=-\left(15\times2\right)-240-6+\left[-\left(36:6\right)\times2\right]\)

\(=\left(-30\right)-240-6+\left[-6\times2\right]\)

\(=\left(-30\right)-240-6+\left(-12\right)\)

\(=-270-6+\left(-12\right)\)

\(=-276+\left(-12\right)\)

\(=-288\)

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)  

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

1) TXĐ: \(D=ℝ\)

 \(9^x+3.6^x=4^{x+1}\)

\(\Leftrightarrow9^x-4.4^x+3.6^x=0\)

\(\Leftrightarrow\dfrac{9^x}{4^x}-4+3.\dfrac{6^x}{4^x}=0\)

\(\Leftrightarrow\left(\dfrac{9}{4}\right)^x+3\left(\dfrac{6}{4}\right)^x-4=0\)

\(\Leftrightarrow\left[\left(\dfrac{3}{2}\right)^2\right]^x+3\left(\dfrac{3}{2}\right)^x-4=0\)

\(\Leftrightarrow\left[\left(\dfrac{3}{2}\right)^x\right]^2+3\left(\dfrac{3}{2}\right)^x-4=0\)

\(\Leftrightarrow\left[\left(\dfrac{3}{2}\right)^x-1\right]\left[\left(\dfrac{3}{2}\right)^x+4\right]=0\)

\(\Leftrightarrow\left(\dfrac{3}{2}\right)^x=1\) (vì \(\left(\dfrac{3}{2}\right)^x>0\))

\(\Leftrightarrow x=0\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{0\right\}\)

2)

a) \(D=ℝ\)

Với \(m=1\) thì (1) thành:

\(\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}+\left(\sqrt{2-\sqrt{3}}\right)^{\left|x\right|}=4\)

Để ý rằng \(\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=1\) \(\Leftrightarrow\sqrt{2-\sqrt{3}}=\dfrac{1}{\sqrt{2+\sqrt{3}}}\)

Do đó pt \(\Leftrightarrow\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}+\left(\dfrac{1}{\sqrt{2+\sqrt{3}}}\right)^{\left|x\right|}-4=0\)

Đặt \(\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}=t\left(t\ge1\right)\) thì pt thành:

\(t+\dfrac{1}{t}-4=0\)

\(\Leftrightarrow t^2-4t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2+\sqrt{3}\left(nhận\right)\\t=2-\sqrt{3}\left(loại\right)\end{matrix}\right.\)

Vậy \(\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}=2+\sqrt{3}\)

\(\Leftrightarrow\left|x\right|=2\)

\(\Leftrightarrow x=\pm2\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{\pm2\right\}\)]

2b) Đặt \(f\left(x\right)=\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}+\left(\sqrt{2-\sqrt{3}}\right)^{\left|x\right|}\)

\(f\left(x\right)=\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}+\dfrac{1}{\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}}\)

Đặt \(\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}=t\left(t\ge1\right)\) thì \(f\left(x\right)=g\left(t\right)=t+\dfrac{1}{t}\)

\(g'\left(t\right)=1-\dfrac{1}{t^2}\ge0,\forall t\ge1\)

Lập BBT, ta thấy để \(g\left(t\right)=4m\) có nghiệm thì \(t\ge1\). Tuy nhiên, với \(t>1\) thì sẽ có 2 số \(x\) thỏa mãn \(\left(\sqrt{2+\sqrt{3}}\right)^{\left|x\right|}=t\) (là \(\log_{\sqrt{2+\sqrt{3}}}t\)

 và \(-\log_{\sqrt{2+\sqrt{3}}}t\))

Với \(t=1\), chỉ có \(x=0\) là thỏa mãn. Như vậy, để pt đã cho có nghiệm duy nhất thì \(t=1\)

\(\Leftrightarrow m=g\left(1\right)=2\)

 Vậy \(m=2\)

Bài 14:

c) 

\(H=\dfrac{\dfrac{3}{7}-\dfrac{3}{17}+\dfrac{3}{37}}{\dfrac{5}{7}-\dfrac{5}{17}+\dfrac{5}{37}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{10}{2}-\dfrac{10}{3}+\dfrac{10}{4}-\dfrac{10}{5}}\\ =\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}}{10\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\right)}\\ =\dfrac{3}{5}+\dfrac{1}{10}\\ =\dfrac{6}{10}+\dfrac{1}{10}\\ =\dfrac{7}{10}\)

\(\left(-18-12\right):5\)

\(=-\left(18+12\right):5\)

\(=-30:5\)

\(=-6\)

   (- 18 - 12) : 5

= -(18 + 12) : 5

= - 30 : 5

=  - 6 

43 - 28 = 15

15