(5x-1)(5x+1)=25x²-7x+15

<=> 25x²-1-25x²+7x-15=0

<=>7x=16

=>x=16/7

Bài 2:

A=4x-x²+3=-(x²-4x+4)+7=-(x-2)²+7

=> A_max=7 khi -(x-2)²=0 => x=2

B=x-x²=-(x²-x+1/4)+1/4=-(x-1/2)²+1/4

=>B_max=1/4 khi -(x-1/2)²=0 => x=1/2

N=2x-2x²-5=2x-2x²-1/2-9/2=-2(x²-x+1/4)-9/2=-2(x-1/2)²-9/4

=>N_max=-9/4 khi -2(x-1/2)²=0 => x=1/2

Gọi số mol của NaOH là x, số mol của KOH là y ta có PTHH:

NaOH + HCl --------> NaCl + H₂O

x x

KOH + HCl --------> KCl + H₂O

y y

Theo đề ra, ta có: 40x+56y=6,08g

58,5x+74,5y=8,3g

=> x=0,04mol y=0,08mol

n_HCl=0,04+0,08=0,12mol

a)m_HCl=0,12.36,5=4,38g

b)m_NaCl=0,04.58,5=2,34g

m_KCl=0,08.74,5=5,96g

1.a)a²-ab+a-b=a²+a-ab-b=a(a+1)-b(a+1)=(a-b)(a+1)

b)x³-2xy-x²y+2y²=x³-x²y-2xy+2y²=x²(x-y)-2y(x-y)=(x²-2y)(x-y)

c)a²-x²+2a+1=a²+2a+1-x²=(a+1)²-x²=(a+1-x)(a+1+x)

d)m²-a²+2ab-b²=m²-(a²-2ab+b²)=m²-(a-b)²=(m-a+b)(m+a-b)

e)25b⁴-x²-4x-4=(5a²)²-(x²+4x+4)=(5a²)²-(x+2)²=(5a²-x-2)(5a²+x+2)

f)3x²+6xy+3y²-3z²=3(x²+2xy+y²-z²)=3((x+y)²-z²)=3(x+y-z)(x+y+z)

g)a²-2ax-b²-2by+x²-y²=a²-2ax+x²-(b²+2by+y²)=(a-x)²-(b+y)²=(a-x-b-y)(a-x+b+y)

\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{x+5-x}{x\left(x+5\right)}=\dfrac{5}{x\left(x+5\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+9}-\dfrac{1}{x+10}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+10}=\dfrac{x+10-x}{x\left(x+10\right)}=\dfrac{10}{x\left(x+10\right)}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

=\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)\)

mà a+b+c=0

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{0}{abc}\right)=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

2x2²x2³.....x2^x=32768=2¹⁵

=>2^1+2+3+...+x=2¹⁵ =>1+2+3+...+x=15

=>\(\dfrac{x\left(x+1\right)}{2}=15=>x\left(x+1\right)=30=5.\left(5+1\right)\)

=>x=5

Q=x²-2x+7=x²-2x+1+6=(x-1)²+6 (x-1)²\(\ge0\)

Biểu thức Q có GTNN là 6 khi (x-1)²=0

=>x-1=0 => x=1

x+y+z=1 => (x+y+z)²=1²=1

(x+y+z)²=x²+y²+z²+2xy+2xz+2yz=1 mà x²+y²+z²=1

=>2xy+2xz+2yz=2(xy+xz+yz)=0

=>xy+yz+xz=0