1. \(2-\sqrt{\left(3x+1\right)^2}=35\)

<=> \(\left|3x+1\right|=-33\) => pt vô nghiệm

2. \(\sqrt{\left(-2x+1\right)^2}+5=12\)

<=> \(\left|1-2x\right|=12-5\)

<=> \(\left|1-2x\right|=7\)

<=> \(\orbr{\begin{cases}1-2x=7\left(đk:x\le\frac{1}{2}\right)\\2x-1=7\left(đk:x>\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}2x=-6\\2x=8\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-3\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

Vậy S = {-3; 4}

3. ĐKXĐ: \(\sqrt{x^2-1}\ge0\) <=> \(x^2-1\ge0\) <=> \(x^2\ge1\) <=> \(\orbr{\begin{cases}x\ge1\\x\le1\end{cases}}\)

\(\sqrt{x^2-1}+4=0\) <=> \(\sqrt{x^2-1}=-4\)

=> pt vô nghiệm

4. Đk: \(\hept{\begin{cases}\sqrt{5x+7}\ge0\\\sqrt{x+3}>0\end{cases}}\) <=> \(\hept{\begin{cases}5x+7\ge0\\x+3>0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge-\frac{7}{5}\\x>-3\end{cases}}\) => x \(\ge\)-7/5

Ta có: \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)

<=> \(\left(\frac{\sqrt{5x+7}}{\sqrt{x+3}}\right)^2=16\)

<=> \(\frac{\left(\sqrt{5x+7}\right)^2}{\left(\sqrt{x+3}\right)^2}=16\)

<=> \(\frac{5x+7}{x+3}=16\)

=> \(5x+7=16\left(x+3\right)\)

<=> \(5x+7=16x+48\)

<=> \(5x-16x=48-7\)

<=> \(-11x=41\)

<=> \(x=-\frac{41}{11}\)ktm

=> pt vô nghiệm

48 + 46 + 50 + 44 - 47 - 43 - 41 - 45

= ( 50 - 47 ) + ( 48 - 45 ) + ( 46 - 43 ) + ( 44 - 41 )

= 3 + 3 + 3 + 3 = 12

\(A=\frac{1}{x^2+y^2}+\frac{501}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+\frac{500}{xy}\)

\(\ge\frac{5}{\left(x+y\right)^2}+\frac{500}{\frac{\left(x+y\right)^2}{2}}=5+1000=1005\)

Dấu "=" xảy ra \(< =>x=y=\frac{1}{2}\)

đoán là sai

\(A=\frac{1}{x^2+y^2}+\frac{501}{xy}\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1001}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1001}{\frac{\left(x+y\right)^2}{2}}\ge4+2002=2006\)

Dấu "=" xảy ra khi  x = y = 1/2

Bài làm 

a) A= 0,1,2,3,4,5,6,7,8,9

b)B= 16,18,20,22

                     hok tốt                 

1) \(x^4+2x^3-9x^2-10x-24\)

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

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

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

2) \(6x^4+7x^3+5x^2-x-2\)

\(=6x^4-3x^3+10x^3-5x^2+10x^2-5x+4x-2\)

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

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

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

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

3) \(2x^4+3x^3+2x^2-1\)

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

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

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

4) \(x^3-x^2-x-2\)

\(=x^3-2x^2+x^2-2x+x-2\)

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

\(\frac{-27}{5}.\frac{-4}{15}+\frac{-4}{15}.\left(2010+\frac{27}{5}\right)\)

\(=\frac{-4}{15}.\left(\frac{-27}{5}+2010+\frac{27}{5}\right)\)

\(=\frac{-4}{15}.2010\)

\(=-536\)

Học tốt

\(-\frac{27}{5}.\left(-\frac{4}{15}\right)+\left(-\frac{4}{15}\right).\left(2010-\frac{-27}{5}\right)\)

\(=\left(-\frac{4}{15}\right)\left(2010-\frac{-27}{5}+\frac{-27}{5}\right)\)

\(=-\frac{4}{15}.\left(2010+\frac{27}{15}-\frac{27}{15}\right)\)

\(=-\frac{4}{15}.2010=-536\)

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a) Lập bảng xét dấu

x              0           1              2

x        -     0      +    |       +      |       +

x - 1 -      |       -     0     +      |      +

x - 2 -    |          -    |     -         |      +

Xét các TH xảy ra

TH1: x \(\le\)0 => pt trở thành: -x - 2(1 - x) + 3(2 - x) = 4

<=> - x - 2 + 2x + 6 - 3x = 4 <=> -2x = 4 - 4 <=> -2x = 0 <=> x = 0 (tm)

TH2: 0 < x \(\le\)1 => pt trở thành: x - 2(1 - x) + 3(2 - x) = 4

<=> x - 2 + 2x + 6 - 3x = 4 <=> 4 = 4 (luôn đúng)

TH3: 1 < x \(\le\)2 => pt trở thành: x - 2(x - 1) + 3(2 - x) = 4

<=> x - 2x + 2 + 6 - 3x = 4 <=> -4x = 4 - 8 <=> -4x = -4 <=> x = 1 (ktm)

TH4: x > 2 => pt trở thành: x - 2(x - 1) + 3(x - 2)  = 4

<=> x - 2x + 2 + 3x - 6 = 4 <=> 2x = 4 + 4 <=> 2x = 8 <=> x = 4 (tm)

Vậy ....