Bạn ơi cả 3x-5  chia cho 12 hay chỉ có 5 chia cho 12 thui z ạ

Để biểu thức P có giá trị là 5 thì :

2x^2+3x = 5

<=> 2x^2+3x-5 = 0

<=> (2x^2-2x)+(5x-5) = 0

<=> (x-1).(2x+5) = 0

<=> x-1=0 hoặc 2x+5=0

<=> x=1 hoặc x=-5/2

Tk mk nha

Ta có \(P=2x^2+3x=5\)

\(\Rightarrow x.\left(2x+3\right)=5\)

Do đó x và 2x + 3 thuộc ước của 5 

Mà \(Ư\left(5\right)\in\left\{1;5;-1;-5\right\}\)

Mặt khác 2x + 3 > x .

Tự làm tiếp :v 

\(\left(3x-\frac{5}{12}\right)^2-\frac{121}{64}=0\)

\(\left(3x-\frac{5}{12}\right)^2=\frac{121}{64}\)

\(3x-\frac{5}{12}=\sqrt{\frac{121}{64}}\)

\(3x-\frac{5}{12}=\frac{11}{8}\)

\(3x=\frac{11}{8}+\frac{5}{12}\)

\(3x=\frac{33}{24}+\frac{10}{24}\)

\(3x=\frac{43}{24}\)

\(x=\frac{43}{24}:3\)

\(x=\frac{43}{24}\cdot\frac{1}{3}\)

\(x=\frac{43}{72}\)

\(\left(3x-\dfrac{5}{12}\right)^2-\dfrac{121}{64}=0\)

\(\left(3x-\dfrac{5}{12}\right)^2\) \(=0+\dfrac{121}{64}\)

\(3x-\dfrac{5}{12}\) \(=\sqrt{\dfrac{121}{64}}=\dfrac{11}{8}\)

\(3x\) \(=\dfrac{11}{8}+\dfrac{5}{12}=\dfrac{33+10}{24}=\dfrac{43}{24}\)

\(x\) \(\dfrac{3.43}{24}=\dfrac{43}{8}\)

a: \(A=\dfrac{19}{5}xy^2\cdot x^3y=\dfrac{19}{5}x^4y^3\)

b: Hệ số là 19/5 và bậc là 7

c: Khi x=1 và y=2 thì \(A=\dfrac{19}{5}\cdot1^4\cdot2^3=\dfrac{19}{5}\cdot8=\dfrac{152}{5}\)

\(b,4x^2-x-5=0\)

\(\Leftrightarrow4x^2-5x+4x-5=0\)

\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)

\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)

Bài 2

\(a,x^3+5x^2+3x-9\)

\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)

\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)

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

b,\(x^3-7x-6\)

\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)

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

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

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

c,\(3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

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

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

\(4x^2-x-5=0\)

<=>  \(4x^2+4x-5x-5=0\)

<=>   \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự lm tiếp

Bài 1:

a)\(28x^3+15x^2+75x+125=0\)

\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)

Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)

\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)

b)\(4x^2-x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)

\(\Rightarrow x=-1;x=\frac{5}{4}\)

Bài 2:

a)\(x^3+5x^2+3x-9\)

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

b)\(x^3-7x-6\)

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

c)\(3x^3-7x^2+17x-5\)

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

\(4x^2-x-5=0\)

<=> \(4x^2+4x-5x-5=0\)

<=>  \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự giải nốt

Bài làm

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

\(\frac{x}{2}=\frac{y}{5}=\frac{y-x}{5-2}=\frac{42}{3}=14\)

Do đó: \(\hept{\begin{cases}\frac{x}{2}=14\\\frac{y}{5}=14\end{cases}\Rightarrow\hept{\begin{cases}x=28\\y=70\end{cases}}}\)

Vậy x = 28; y = 70

# Học tốt #

\(\frac{x}{2}=\frac{y}{5}\)=>5x=2y

=>2y-5x=0

=>2y-2x-3x=0

=>2(y-x)=3x

=>2.42=3x

=>3x=84=>x=26,y=68

nhầm x=28,y=70

khó quá mình bó tay