\(-\left|15\right|\times3-\left(2-5\right)^2+\left(-1005\right)^5\)

\(\Leftrightarrow-15\times3-3^2-1005^5\)

\(\Leftrightarrow-45+9-1005^5\)

\(\Leftrightarrow-36-1005^5\)

LẤY A=100:2+1=51(SỐ)

B=(102-2):2+1=51(SỐ)

C=(1005-5):5+1=201(SỐ)

NHỚ **** CAC BẠN

Áp dụng công thức:

Số số hạng = (số cuối - số đầu) : khoảng cách + 1

Tổng = (số cuối + số đầu) x số số hạng : 2

USE máy tính

a) \(\left(-5\right)\left(x-2\right)^2+360=\left(-150\right)\cdot3+43\cdot5\)

\(-5\cdot\left(x-2\right)^2+360=-235\)

\(-5\cdot\left(x-2\right)^2=-595\)

\(\left(x-2\right)^2=119\)

\(\left(x-2\right)^2=\left(\pm\sqrt{199}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{119}\\x-2=-\sqrt{119}\end{cases}\Rightarrow\orbr{\begin{cases}x=\sqrt{119}+2\\x=2-\sqrt{119}\end{cases}}}\)

b) \(\left(x+5\right)-\left(3x+9\right)=-16\)

\(x+5-3x-9=-16\)

\(-2x-4=-16\)

\(-2x=-12\)

\(x=6\)

c) \(3\left(x+2\right)-\left(15-x\right)\cdot6=160+\left(-1\right)^{1005}\)

\(3x+6-90+6x=160-1\)

\(9x-84=159\)

\(9x=243\)

\(x=27\)

d) \(x\left(x-1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\left\{0;1\right\}\\x=\left\{2;-2\right\}\end{cases}}}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}\)

\(=\frac{2^{1005}.7^{1005}.5^{1006}}{2^{1007}.5^{1004}.7^{1004}}\)

\(=\frac{5^2.7}{2^2}=\frac{25.7}{4}=\frac{175}{4}\)

TRẦN TIỂU HY ƠI, BẠN TRÌNH BÀY RA GIÙM MK NHA. MK KO HIỂU LẮM