\(x^4=6,25^2\)

\(\Leftrightarrow x^4=2,5^4\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2,5\\x=2,5\left(loại\right)\end{array}\right.\)

Số gạo tẻ cân nặng là:

40 x 30 = 1200 (kg)

Số gạo nếp cân nặng là:

20 x 40 = 800 (kg)

Xe đó chở số kg gạo là:

1200 + 800 = 2000 (kg) 

Đổi 2000 kg = 2 tấn

                  Đáp số: 2 tấn

Bài 1:

b) \(16x^2-8x+1=\left(4x-1\right)^2\)

c) \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left[\left(x+3\right)\left(x+6\right)\right]\left[\left(x+4\right)\left(x+5\right)\right]+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

Đật \(x^2+9x+19=t\) , pt trở thành

\(\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(x^2+9x+19\right)^2\)

d) \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

e) \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

a)_ Sai đề

lm dưới rùi

\(B=\left|x-2006\right|-\left|2007-x\right|=\left|x-2006\right|-\left|x-2007\right|\)

Áp dụng bđt: \(\left|A\right|-\left|B\right|\le\left|A-B\right|\)

=>\(B\le\left|x-2006-x+2007\right|=1\)

Vậy GTLN của B là 1 khi \(2006\le x\le2007\)

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

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

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\) 

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{array}\right.\)

chắc trả lời sai

Hạo ơiiiiiiiii

mẹ cx muốn

Bài 1:

a) \(25x^2+3-10x=\left(25x^2-10x+1\right)+2=\left(5x-1\right)^2+2>0\)

=>đpcm

b) \(-9x^2-2+6x=-\left(9x^2-6x+1\right)-1=-\left(3x-1\right)^2-1< 0\)

=>đpcm

Bài 2:

\(A=4x^2+3-4x=\left(4x^2-4x+1\right)+2=\left(2x-1\right)^2+2\ge2\)

Vậy \(x=\frac{1}{2}\) thì A đạt GTNN là 2

\(B=-x^2+10x-28=-\left(x^2-10x+25\right)-3=-\left(x-5\right)^2-3\le-3\)

Vậy x=5 thì B đạt GTLN là -3

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