\(P=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)

\(P=\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right):\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)\)

\(P=\frac{12}{10}:\frac{11}{12}\)

\(P=\frac{6}{5}\times\frac{12}{11}\)

\(P=\frac{72}{55}\)

\(P=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)

\(\Rightarrow P=\frac{\frac{3}{2}-\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}-\frac{2}{3}+\frac{1}{12}}\)\(\Rightarrow P=\frac{\frac{15}{10}-\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}-\frac{8}{12}+\frac{1}{12}}\)

\(\Rightarrow P=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

Vì \(\hept{\begin{cases}a+b+c\ne0\\\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\text{ mà }a = 2012\Rightarrow\hept{\begin{cases}b=2012\\c=2012\end{cases}}\)

\(\frac{5^{100}.9^{1000}}{3^{2000}.25^{50}}\)

\(=\frac{5^{100}.\left(3^2\right)^{1000}}{3^{2000}.\left(5^2\right)^{50}}\)

\(=\frac{5^{100}.3^{2000}}{3^{2000}.5^{100}}\)

\(=1\)

Vì \(\frac{x}{4}=\frac{4}{7}\)\(\Rightarrow x=\frac{4}{7}.4=\frac{16}{7}\)

Mà xy = 12

\(\Rightarrow\frac{16}{7}.y=12\)

\(\Rightarrow y=\frac{21}{4}\)

\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)=\(\frac{100^4}{100^5}\)=\(\frac{1}{100}\)
 

\(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)

\(=\frac{\left(-7\right)^n}{\left(-7\right)^n:\left(-7\right)}\)

\(=\frac{\left(-7\right)^n}{\left(-7\right)^n.\frac{1}{\left(-7\right)}}\)

\(=\frac{1}{\frac{1}{-7}}\)

\(=-7\)

-7 nha bạn

Đáp án: P≥12√407P≥12407

Giải thích các bước giải:

Giả sử PP đạt giá trị nhỏ nhất khi a=x,b=y,c=za=x,b=y,c=z

Ta có:

2(a3+a3+x3)≥2⋅33√a3⋅a3⋅x3=6xa22(a3+a3+x3)≥2⋅3a3⋅a3⋅x33=6xa2

3(b3+b3+y3)≥3⋅33√b3⋅b3⋅y3=9yb23(b3+b3+y3)≥3⋅3b3⋅b3⋅y33=9yb2

4(c3+c3+z3)≥4⋅33√c3⋅c3⋅z3=12zc24(c3+c3+z3)≥4⋅3c3⋅c3⋅z33=12zc2

Cộng vế với vế của 33 bất đẳng thức trên

→2(2a3+3b3+4c3)+2x3+3y3+4z3≥6xa2+9yb2+12zc2→2(2a3+3b3+4c3)+2x3+3y3+4z3≥6xa2+9yb2+12zc2

→2P+2x3+3y3+4z3≥6xa2+9yb2+12zc2→2P+2x3+3y3+4z3≥6xa2+9yb2+12zc2

Chọn x,y,zx,y,z thỏa mãn:

⎧⎨⎩6x1=9y2=12z3x2+2y2+3z2=1{6x1=9y2=12z3x2+2y2+3z2=1

→⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x=6√407y=8√407z=9√407→{x=6407y=8407z=9407

→P≥12√407→P≥12407 khi a=6√407,b=8√407,c=9√407

TL

( 0,25 )⁴ . 1024=4

Hoktot~

( 0,25 )⁴ . 1024

= \(\left(\frac{1}{4}\right)^4.2^{10}\)

\(=\left(\frac{1}{2^2}\right)^4.2^8.2^2\)

\(=\frac{1}{2^8}.2^8.4\)

\(=4\)