​Taco:-3x²>hoặc=0

         5x>hoặc=0

=)-3x²​-5x+7luôn >hoặc =7

Vậymax của đa thức đó là 7

2.

a) Vì \(\left|2x+1\right|\ge0\forall x\in R\\ \Rightarrow3\left|2x+1\right|\ge0\forall x\in R\\ \Rightarrow3\left|2x+1\right|-4\ge-4\forall x\in R\\ \Rightarrow A\ge-4\forall x\in R\)

Vậy GTNN của A là -4 đạt được khi \(x=-\dfrac{1}{2}\)

Mai mk phải nộp rồi ! Các bn ơi giúp mk với! Help Me ! Thank you !

a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)

=>121-8x=9x+207

=>-17x=86

hay x=-86/17

b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)

=>|2x-1|=9/5

=>2x-1=9/5 hoặc 2x-1=-9/5

=>2x=14/5 hoặc 2x=-4/5

=>x=7/5 hoặc x=-2/5

mn ơi giúp nhé

Bài 1:

a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

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

c) \(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)

d) \(4x^2+9x+5=4x^2+4x+5x+5=4x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(4x+5\right)\)

Bài 2:

không rõ đề --> k lm

bai 2 la tim x de cac bieu thuc sau duong

a) Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{3x+2-3x+1}{5x+7-5x-1}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5x+7}{2}\\3x-1=\dfrac{5x+1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4=5x+7\\6x-2=5x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)(Nhận)

Vậy x = 3

b) Giống vậy :)

a, \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)

\(\Leftrightarrow15x^2+13x=15x^2+16x-9\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b, Tương tự

1: \(\Leftrightarrow\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

a/ đk: x khác -7/5 ; x khác -1/5

pt <=> \(\dfrac{\left(3x+2\right)\left(5x+1\right)}{\left(5x+7\right)\left(5x+1\right)}=\dfrac{\left(3x-1\right)\left(5x+7\right)}{\left(5x+7\right)\left(5x+1\right)}\)

\(\Rightarrow15x^2+13x+2=15x^2+16x-7\)

\(\Leftrightarrow15x^2+13x-15x^2-16x^2=-7-2\)

\(\Leftrightarrow-3x=-9\Leftrightarrow x=3\left(tm\right)\)

vậy x = 3

b/ đk: x khác -1/2; x khác -3

pt <=> \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(2x+1\right)\left(x+3\right)}=\dfrac{\left(0,5x+2\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+3\right)}\)

\(\Rightarrow x^2+4x+3=x^2+4,5x+2\)

\(\Leftrightarrow x^2+4x-x^2-4,5x=2-3\)

\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\left(tm\right)\)

vậy x = 2

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\Rightarrow2\left(3x+2\right)=5x+7\)

\(\Rightarrow6x+4=5x+7\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b) Ta có: \(\dfrac{0,5x+2}{x+3}=\dfrac{2\left(0,5x+2\right)}{2\left(x+3\right)}=\dfrac{x+4}{2x+6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}=\dfrac{x+4}{2x+6}=\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(2x+6\right)-\left(2x+1\right)}=\dfrac{3}{5}\)

\(\Rightarrow5\left(x+1\right)=3\left(2x+1\right)\)

\(\Rightarrow5x+5=6x+3\)

\(\Leftrightarrow x=2\)