Bài 1:

a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)

\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)

\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

đa thức lớp 5 hả bạm

đa thức lớp 5 à

mình ghi sao đề, các bạn ko cần làm đâu

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

\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)

\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)

\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)

\(\Rightarrow4x=21\)

\(\Rightarrow x=\dfrac{21}{5}\)

1,a, \(\left(2x+1\right)\left(4x^2-2x+1\right)-8x\left(x^2+2\right)=17\)

\(\Leftrightarrow8x^3+1-8x^3-16x=17\)

\(\Leftrightarrow-16x=16\)

\(\Leftrightarrow x=-1\)

\(b,x^2-2x+5\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)

2,\(M=x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5\ge5\)

Dấu "=" xảy ra <=> x +  1 = 0

                        <=> x = -1

Vậy \(M_{min}=5\Leftrightarrow x=-1\)

a) \(3\frac{1}{3}\left(3\frac{1}{4}+2x\right)=6\frac{2}{3}\)

   \(3\frac{1}{3}\times3\frac{1}{4}+2x=6\frac{2}{3}\)

   \(10\frac{5}{6}+2x=6\frac{2}{3}\)

   \(2\times x=6\frac{2}{3}+10\frac{5}{6}=17,5\)

   \(x=17,5\div2=8,75\)

Vậy x = 8,75

b) \(x-25\%x=\frac{6}{11}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)

   \(x-\frac{25}{100}x=\frac{6}{11}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\)

   \(x-\frac{1}{4}\times x=\frac{6}{11}\times1\frac{7}{12}=\frac{19}{22}\)

   \(x\times x=\frac{19}{22}+\frac{1}{4}=\frac{49}{44}\)

\(\Rightarrow2x\left(x\times x\right)=\frac{49}{44}\)   

   \(x=\frac{49}{44}\div2=\frac{49}{88}\)

   Vậy x = \(\frac{49}{88}\)

c) \(\left(4,5-2x\right)\times1\frac{4}{7}=\frac{11}{14}\)

   \(4,5-2x\times1\frac{4}{7}=\frac{11}{14}\)

   \(-2x\times1\frac{4}{7}=\frac{11}{14}-4,5=-3\frac{5}{7}\)

   \(-2\times x=-3\frac{5}{7}\div1\frac{4}{7}=-2\frac{4}{11}\)

   \(x=-2\frac{4}{11}\div\left(-2\right)=1\frac{2}{11}\)

Vậy x = \(1\frac{2}{11}\)

d) \(-3^2-|2x+3|=4\)

    \(9-|2x+3|=4\)

   \(-|2x+3|=4-9=-5\)

   \(-|2x|=-5-|3|=-8\)

   \(-|x|=-8\div2=-4\)

    \(-x=4\Rightarrow x=-4\)

   Vậy x = -4 (-x được xem là số đối của x)

1) Tìm x

a) x + 12 = 10 + 36

  x + 12 = 46

  x = 46 - 12

  x = 34

b) 23 + x = 2 * 3

  23 + x = 6

  x = 6 - 23

  x = -17

2) Tính nhanh

a)10 * 5 + 20 * 5 = 5 * (10 + 20) = 5 * 30 = 150

b) 6 * 2 + 4 * 2 + 10 * 2 = 2 * (6 + 4 + 10) = 2 * 20 = 40

x + 12 = 10 + 36

x + 12 = 46

x = 36

23 + x = 2 x 3

23 + x = 6

x = -17

x + 12 = 10 + 36

x +12 = 46

x = 46 -12

x = 34 

23 + x = 2 x 3 

23 + x = 6

x = -17

tính nhanh tự làm nha

a.

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

b.

\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\)

\(15x^2+30=0\\ \Rightarrow x^2+2=0\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

\(\left(2x-1\right)^2.4=1\\ \Rightarrow\left(2x-1\right)^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{1}{2}\\2x-1=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)