\(\text{a)}3x^2-2x\left(5+1.5x\right)+10\\ =3x^2-10x+3x^2+10\\ =\left(3x^2+3x^2\right)-10x+10\\ =6x^2-10x+10\)

\(\text{b)}5x\left(3x^2-12x+6\right)+4x^3\\ =15x^3-60x^2+30x+4x^3\\ =\left(15x^3+4x^3\right)-60x^2+30x\\ =19x^3-60x^2+30x\)

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Sàm

P NGUYÊN TỐ LỚN HƠN 3 NÊN P LẺ SUY RA 5P LẺ NÊN 5P+1 CHIA HẾT CHO 2.P NGUYÊN TỐ LỚN HƠN 3 P CÓ DẠNG 3K+1 HOẶC 3K+2.

P= 3K+2 SUY RA 10P+1 CHIA HẾT CHO 3(LOẠI) VẬY P =3K+1 SUY RA 5P+1 CHIA HẾT CHO 3 MÀ (3,2)=1 SUY RA 5P+1 CHIA HẾT CHO 6

tick đúng nhé bạn

\(A=\dfrac{4\cdot5^{10}\cdot5^{10}}{75^{10}}=\dfrac{4\cdot5^{20}}{\left(3\cdot25\right)^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot25^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot\left(5^2\right)^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot5^{20}}=\dfrac{4}{3^{10}}\)

\(B=\dfrac{\left(0.8\right)^5}{\left(0.4\right)^6}=\dfrac{\left(\dfrac{4}{5}\right)^5}{\left(\dfrac{2}{5}\right)^6}=\dfrac{\left(2\cdot\dfrac{2}{5}\right)^5}{\left(\dfrac{2}{5}\right)^6}=\dfrac{2^5\cdot\left(\dfrac{2}{5}\right)^5}{\left(\dfrac{2}{5}\right)^5\cdot\dfrac{2}{5}}=\dfrac{2^5}{\dfrac{2}{5}}=2^5\cdot\dfrac{5}{2}=\dfrac{32\cdot5}{2}=80\)

\(C=\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\dfrac{3^8}{3^6}=3^2=9\)

\(D=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{^{20}}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=226\)

Ta có nghiệm của đa thức thỏa mãn :

\(x^2-4x=0\\ \Leftrightarrow x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy nghiệm của đa thức là \(x=0\) hoặc \(x=4\)

\(\text{a)}\) \(\left(a+b+c\right)^2\)

\(=\left(a+b+c\right)\left(a+b+c\right)\)

\(=a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)\)

\(=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2\)

\(=a^2+b^2+c^2+\left(ab+ba\right)+\left(ac+ca\right)+\left(bc+cb\right)\)

\(=a^2+b^2+c^2+2ab+2ac+2bc\)

\(\text{b)}\) \(\left(a-b+c\right)^2\)

\(=\left(a-b+c\right)\left(a-b+c\right)\)

\(=a\left(a-b+c\right)-b\left(a-b+c\right)+c\left(a-b+c\right)\)

\(=a^2-ab+ac-ba-b^2+bc+ca-cb+c^2\)

\(=a^2-b^2+c^2-\left(ab+ba\right)+\left(ac+ac\right)+\left(bc-bc\right)\)

\(=a^2-b^2+c^2-2ab+2ac\)

\(\text{Câu 1 :}\)

\(\text{a)}\) \(2^{225}\) \(\text{và}\) \(3^{150}\)

\(\text{Ta có :}\)\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

\(\text{Mà}\) \(8^{75}< 9^{75}\)

\(\Rightarrow2^{225}< 3^{150}\)

\(\text{Vậy}\) \(2^{225}< 3^{150}\)

\(\text{b)}\) \(2^{91}\) \(\text{và}\) \(5^{35}\)

\(\text{Ta có :}\) \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

\(\text{Mà}\) \(8192^7>3125^7\) \(\)

\(\Rightarrow2^{91}>5^{35}\)

\(\text{Vậy}\) \(2^{91}>5^{35}\)

\(\text{c)}\) \(3^{4000}\) \(\text{và}\) \(9^{2000}\)

\(\text{Ta có :}\) \(9^{2000}=\left(3^2\right)^{2000}=3^{4000}\)

\(Mà\) \(3^{4000}=3^{4000}\)

\(\text{Vậy}\) \(3^{4000}=9^{2000}\)

\(\text{d)}\) \(2^{332}\) \(\text{và}\) \(3^{223}\)

\(\text{Ta có : }\) \(2^{332}< 2^{333}\)

\(3^{223}>3^{222}\) \(\left(1\right)\)

\(\text{Ta lại có : }\) \(2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{222}=\left(3^2\right)^{111}=9^{111}\)

\(\text{Mà}\) \(8^{111}< 9^{111}\)

\(\Rightarrow2^{333}< 3^{222}\) \(\left(2\right)\)

\(\text{Từ}\) \(\left(1\right)\) \(\text{và}\) \(\left(2\right)\) \(\text{suy ra : }\)

\(2^{322}< 2^{333}< 3^{222}< 3^{223}\)

\(\Rightarrow2^{332}< 3^{223}\)

\(\text{Vậy}\) \(2^{332}< 3^{223}\)

\(\text{Câu 2 :}\)

\(\text{a)}\) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=27\)

\(\text{b)}\) \(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(9\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot25\right)^{15}}\)\(=\dfrac{9^{10}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot25^{15}}\)\(=\dfrac{\left(3^2\right)^{10}\cdot5^{30}}{3^{15}\cdot\left(5^2\right)^{15}}\)\(=\dfrac{3^{20}\cdot5^{30}}{3^{15}\cdot5^{30}}\)\(=\dfrac{3^{20}}{3^{15}}\)\(=3^4=81\)

\(\text{c)}\) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)

Gọi số vở là a

Ta có a chia 12 dư 2 \(\Rightarrow\) a + 10 chia hết cho 12

         a chia 18 dư 8 \(\Rightarrow\) a + 10 chia hết cho 18

         a chia hết cho 10 \(\Rightarrow\) a + 10 chia hết cho 10

Suy ra a + 10 là BC(12;18;10)

12 = 2².3

18 = 2  .3²

10 = 2  .     5

\(\Rightarrow\) BCNN(12;18;10) = 2².3².5 = 180

\(\Rightarrow\) a + 10 \(\in\) BC(12;18;10) = B(180) = {0;180;360;540;...}

\(\Rightarrow a\in\left\{170;350;530;...\right\}\)

mà 300 < a < 500 nên a = 350

Vậy số sách là 350 quyển

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\\ \left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=1\\\dfrac{3}{4}x=\dfrac{-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\div\dfrac{2}{3}\\x=-\dfrac{1}{2}\div\dfrac{3}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\text{Vậy }\) \(x=\dfrac{3}{2}\) \(\text{hoặc}\) \(x=-\dfrac{2}{3}\)