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\(\left[\left(46-32\right)^2-\left(54-42\right)^2\right]\times36-1872\)

\(=\left(14^2-12^2\right)\times36-1872\)

\(=\left(196-144\right)\times36-1872\)

\(=52\times36-1872\)

\(=1872-1872\)

\(=0\)

;((

\(Ư\left(144\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\right\}\)

\(Ư\left(324\right)=\left\{1,2,3,4,6,9,12,18,27,36,54,81,108,162,324\right\}\)

\(Ư\left(576\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,32,36,48,64,72,96,144,192,288,576\right\}\)

\(Ư\left(1024\right)=\left\{1,2,4,8,16,32,64,128,256,512,1024\right\}\)

\(Ư\left(1296\right)=\left\{1,2,3,4,6,8,9,12,16,24,27,36,48,54,72,81,108,144,162,216,324,432,648,1296\right\}\)

\(Ư\left(2304\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,32,48,64,72,96,128,144,192,256,288,384,576,768,1152,2304\right\}\)

\(Ư\left(2500\right)=\left\{1,2,4,5,10,20,25,50,100,125,250,500,625,1250,2500\right\}\)

\(Ư\left(10000\right)=\left\{1,2,4,5,8,10,16,20,25,40,80,100,125,200,250,400,500,625,1000,1250,2000,2500,5000,10000\right\}\)

\(\left[\left(46-32\right)^2-\left(54-42\right)^2\right]\cdot36-1872\)

\(=\left[\left(14\right)^2-12^2\right]\cdot36-1872\)

\(=\left(196-144\right)\cdot36-1872\)

\(=52\cdot36-1872\)

\(=1872-1872\)

\(=0\)

Bạn dựa vào công thức

a² - n² = (a + n)(a - n)

\(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

\(\Leftrightarrow\left(x^2+11x+12\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

\(\Leftrightarrow\left(x^2+11x+12\right)\left(x^2+11x+28\right)\left(x^2+11x+30\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)

Đặt \(x^2+11x+30=a\)

\(\Leftrightarrow\left(a-18\right)\left(a-2\right)a=36a\left(a+1\right)\)

\(\Leftrightarrow a^3-56a^2=0\)

\(\Leftrightarrow a^2\left(a-56\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=56\end{matrix}\right.\)

Với \(a=0\Leftrightarrow x^2+11x+30=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Với \(a=56\Leftrightarrow x^2+11x+30=56\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)

đề sai rồi bạn , phải là ( x²+11x + 12)(x²+9x+20 ) = 36(x²+11x+30)(x²+11x+31)

a) 23 + 45 = 68

b) \(\left( { - 42} \right) + \left( { - 54} \right) =  - \left( {42 + 54} \right) =  - 96\)

c) \(2025 + \left( { - 2025} \right) = 0\) vì 2025 và \( - 2025\) là 2 số đối nhau.

d) \(15 + \left( { - 14} \right) = 15 - 14 = 1\);

e) \(35 + \left( { - 135} \right) =  - \left( {135 - 35} \right) =  - 100\)

Bài 1:

a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)

\(=12+\left\{45-\left[36:3^2\right]\right\}\)

\(=12+\left\{45-\left[36:9\right]\right\}\)

\(=12+\left\{45-4\right\}\)

\(=12+41\)

\(=53\)

b.\(24:\left[48-\left(42:7\right)^2\right]\)

\(=24:\left[48-6^2\right]\)

\(=24:\left[48-36\right]\)

\(=24:12=2\)

Bài 2 : 

C1 :  { 0;1;2;3;4;5;6;7;8;9;10}

C2 : \(\left\{x\in N|x\le10\right\}\)

pha ngoac ra la dc

con bai 2 C1 {1:2:3:4:5:6:7:8:9:0}

C2 {nt huoc N /n<10}

a) 12 + {45 - [36 : (12 - 9)² ]}

= 12 + [45 - (36 : 3²)]

= 12 + [45 - (36 : 9)]

= 12 + (45 - 4)

= 12 + 41

= 53

a) \(\left(3x-2\right)^2-\left(3x-5\right)\left(3x+2\right)=11\)
\(\Leftrightarrow\left(9x^2-12x+4\right)-\left(9x^2+6x-15x-10\right)=11\)
\(\Leftrightarrow9x^2-12x+4-9x^2-6x+15x+10=11\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)

b) \(\left(4x-3\right)^2-\left(4x-5\right)\left(4x+5\right)=32\)
\(\Leftrightarrow\left(16x^2-24x+9\right)-\left(16x^2-25\right)=32\)
\(\Leftrightarrow16x^2-24x+9-16x^2+25=32\)
\(\Leftrightarrow-24x+2=0\)
\(\Leftrightarrow-24x=-2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy \(S=\left\{\dfrac{1}{12}\right\}\)

c) \(\left(5x-2\right)^2-\left(5x+3\right)\left(5x-5\right)=1\)
\(\Leftrightarrow\left(25x^2-20x+4\right)-\left(25x^2-25x+15x-15\right)=1\)
\(\Leftrightarrow25x^2-20x+4-25x^2+25x-15x+15=1\)
\(\Leftrightarrow-10x+18=0\)
\(\Leftrightarrow-10x=-18\)
\(\Leftrightarrow x=\dfrac{9}{5}\)
Vậy \(S=\left\{\dfrac{9}{5}\right\}\)

d) \(\left(x-4\right)^2-\left(x-7\right)\left(2x-3\right)=5-x^2\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(2x^2-3x-14x+21\right)=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21-5+x^2=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy \(S=\left\{\dfrac{10}{9}\right\}\)