\(\frac{x+73}{27}+1+\frac{x+74}{26}+1=\frac{x+27}{73}+1+\frac{x+26}{74}+1\)

\(\frac{x+73+27}{27}+\frac{x+74+26}{26}=\frac{x+27+73}{73}+\frac{x+26+74}{74}\)

\(\frac{x+100}{27}+\frac{x+100}{26}=\frac{x+100}{73}+\frac{x+100}{74}\)

\(\left(x+100\right)\left(\frac{1}{27}+\frac{1}{26}-\frac{1}{73}-\frac{1}{74}\right)=0\)

mà \(\frac{1}{27}+\frac{1}{26}-\frac{1}{73}-\frac{1}{74}\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

\(C=\frac{97^3+83^3}{180}-97.83\)

\(C=\frac{\left(97+83\right)\left(97^2-97.83+83^2\right)}{180}-97.83\)

\(C=\frac{180\left(97^2-97.83+83^2\right)}{180}-97.83\)

\(C=97^2-97.83+83^2-97.83\)

\(C=97^2-2.97.83+83^2\)

\(C=\left(97-83\right)^2=14^2=196\)

A= 101² = 10201

B= 109² = 11881

C= 34² + 66² + 68*66 = 34² + 2*34*66 + 66² = (34 + 36)² =70² = 4900

D=74² + 24² - 48 *74 = 74² - 2*24 *74 + 24² = (74 - 24)²= 50² = 2500

Lời giải:

Áp dụng HĐT đáng nhớ:

\(\frac{97^3+83^3}{180}-97.83=\frac{(97+83)(97^2-97.83+83^2)}{180}-97.83\)

\(=\frac{180(97^2-97.83+83^2)}{180}-97.83=97^2-97.83+83^2-97.83\)

\(=97^2-2.97.83+83^2\)

\(=(97-83)^2=14^2=196\)

\(A=\dfrac{97^3+83^3}{180}-97.83\)

\(A=\dfrac{\left(97+83\right)\left(97^2-97.83+83^2\right)}{180}-97.83\)

\(A=\dfrac{180\left(97^2-97.83+83^2\right)}{180}-97.83\)

\(A=97^2-97.83+83^2-97.83\)

\(A=97^2-2.97.83+83^2\)

\(A=\left(97-83\right)^2\)

\(A=14^2\)

\(A=196\)

Bài giải:

a) 34² + 66² + 68 . 66 = 34² + 2 . 34 . 66 + 66² = (34 + 66)² = 100² = 10000.

b) 74² + 24² – 48 . 74 = 74² - 2 . 74 . 24 + 24² = (74 - 24)²

=50² =2500

a) Ta có :

\(34^2+2.34.66+66^2=\left(34+66\right)^2=100^2=10000\)

P/S : Áp dụng hằng đẳng thức: \(\left(a+b\right)^2\)

b) \(74^2+24^2-48.74=24^2-2.24.74+74^2=\left(24-74\right)^2=\left(-50\right)^2=2500\)

\(A=\frac{x\left(x-6\right)+74}{13}=\frac{x^2-6x+74}{13}=\frac{\left(x^2-6x+9\right)+65}{13}=\frac{\left(x-3\right)^2}{13}+\frac{65}{13}\ge\frac{65}{13}\)

Dấu "=" xảy ra tại \(x=3\)

\(A=\frac{x\left(x-6\right)+74}{13}=\frac{x^2-6x+9+65}{13}=\frac{\left(x-3\right)^2+65}{13}=\frac{\left(x-3\right)^2}{13}+5\)

Vì \(\left(x-3\right)^2\ge0\)\(\Rightarrow\frac{\left(x-3\right)^2}{13}\ge0\)\(\Rightarrow A\ge5\)

Dấu bằng xảy ra khi \(x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(minA=5\Leftrightarrow x=3\)