thu gọn 7^3*7^5

Oh no nhiều kí tự đặc biệt quá

dễ quá mình ko làm  đc 

\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)

\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)

\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)

1a7 doanle bao anh 1234231 1234231 1234231

\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)

\(\dfrac{1}{2}\times\dfrac{3}{4}:\dfrac{6}{5}=\dfrac{3}{8}:\dfrac{6}{5}=\dfrac{5}{16}\)

\(\dfrac{2}{3}+\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14}{21}+\dfrac{6}{21}=\dfrac{20}{21}\)

\(8\times\dfrac{3}{5}:\dfrac{12}{5}=\dfrac{24}{5}:\dfrac{12}{5}=2\)

\(\dfrac{4}{9}+\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{16}{36}+\dfrac{27}{36}-\dfrac{12}{36}=\dfrac{31}{36}\)

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

\(A_{min}=-1\) khi \(x=-1\)

Câu B chỉ có max, ko có min

\(B=-x^2+3x+4=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{25}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

\(B_{max}=\dfrac{25}{4}\) khi \(x=\dfrac{3}{2}\)

Câu C cũng chỉ có max, không có min

\(C=-4x^2+8x=-4\left(x^2-2x+1\right)+4=-4\left(x-1\right)^2+4\le4\)

\(C_{max}=4\) khi \(x=1\)

Câu D cũng chỉ có max, không có min

\(D=\dfrac{3}{4x^2-4x+1+4}=\dfrac{3}{\left(2x-1\right)^2+4}\le\dfrac{3}{4}\)

\(C_{max}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

(4 câu có 3 câu sai đề)

nhìu dữ

a)3/2

b)-1/3

c)-5/6

d)0

e)-1/2

Bài 2

a=3

b=1/2

c=-1/3

d=0

e=9

f=-2/3

mk ko làm rõ đâu  nhe

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

Dấu \("="\Leftrightarrow x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

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

Dấu \("="\Leftrightarrow x=1\)

\(D=\dfrac{1}{-\left(x^2+2x+1\right)+6}=\dfrac{1}{-\left(x+1\right)^2+6}\ge\dfrac{1}{6}\)

Dấu \("="\Leftrightarrow x=-1\)

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

\(A_{min}=-7\) khi \(x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{3}{2}\)

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

\(C_{min}=-4\) khi \(x=1\)

Biểu thức D không tồn tại cả max lẫn min

1.

$A=2x^2-8x+1=2(x^2-4x+4)-7=2(x-2)^2-7$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A\geq 2.0-7=-7$

Vậy $A_{\min}=-7$ khi $x-2=0\Leftrightarrow x=2$

2.

$B=x^2+3x+2=(x^2+3x+1,5^2)-0,25=(x+1,5)^2-0,25\geq 0-0,25=-0,25$

Vậy $B_{\min}=-0,25$ khi $x=-1,5$

3.

$C=4x^2-8x=(4x^2-8x+4)-4=(2x-2)^2-4\geq 0-4=-4$

Vậy $C_{\min}=-4$ khi $2x-2=0\Leftrightarrow x=1$

4. Để $D_{\min}$ thì $5-x^2-2x$ là số thực âm lớn nhất

Mà không tồn tại số thực âm lớn nhất nên không tồn tại $x$ để $D_{\min}$