Có:\(3\left(x+3\right)\left(x-7\right)+\left(x+4\right)^2+48\)

\(=3\left(x^2-7x+3x-21\right)+\left(x^2+8x+16\right)+48\)

\(=3x^2-21x+9x-63+x^2+8x+16+48\)

\(=4x^2-4x+1\)

\(=\left(2x+1\right)^2\)

Với x=0,5 ta co:\(\left(2x+1\right)^2=\left(2\cdot0,5+1\right)^2=\left(1+1\right)^2=4\)

a)\(x^5+x+1\)

\(=\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

=\(x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

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

\(\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

b)\(x^7+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

vuông cân mới làm được bạn chép sai đề r

a)\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-1=0\\2x-\frac{1}{3}=0\end{array}\right.\) 

\(\Leftrightarrow\left[\begin{array}{nghiempt}5x=1\\2x=\frac{1}{3}\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{1}{6}\end{array}\right.\)

\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\Leftrightarrow\)\(\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)

\(\Leftrightarrow\)\(-x=\frac{2}{7}-\frac{3}{35}+\frac{3}{5}\)

\(\Leftrightarrow-x=\frac{4}{5}\)

\(\Leftrightarrow x=-\frac{4}{5}\)

\(\left(7x+2\right)^{-1}=3^{-2}\)

\(\Leftrightarrow\)\(\frac{1}{7x+2}=\frac{1}{9}\)

\(\Leftrightarrow7x+2=9\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

\(A=5+3\left(2x-1\right)^2\)

Vì \(\left(2x-1\right)^2\ge0\) với mọi x

=>\(5+\left(2x-1\right)^2\ge5\)

Vậy GTNN của A là 5 khi x=1/2

a) Xét ΔHAC và ΔKBC có:

       \(\widehat{AHC}=\widehat{BKC}=90\left(gt\right)\)

       \(\widehat{C}\) : góc chung

=>ΔHAC~ΔKBC(g.g)

b)Vì ΔHAC~ΔKBC(cmt)

=>\(\frac{HC}{AC}=\frac{KC}{BC}\) hay \(\frac{AC}{HC}=\frac{BC}{KC}\)

Xét ΔABC và ΔHKC có:

       \(\widehat{C}\) : góc chung

    \(\frac{AC}{HC}=\frac{BC}{KC}\) (cmt)

=>ΔABC~ΔHKC(c.g.c)

c)Vì ΔABC~ΔHKC(cmt)

=>\(\widehat{ABC}=\widehat{HKC}=50\)

\(x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)

Thể tích bể chứa nước đó là:

               1, 8 x 1,5 x 1 = 2,7 (m3)

Đổi: 2,7 m3 = 2700 dm3 = 2700 l

Lượng nước cần phải đổ thêm vào bể là:

               2700 - 800 = 1900 (l)

                         Đáp số: 1900 l nước