\(x=7;y=7\)

7

a) 3/4x = 1

         x = 1 : 3/4

         x = 4/3

b) 4/7x = 9/8 - 0,125

    4/7x = 1

        x = 1 : 4/7

        x = 7/4

a)\(\frac{3}{4}\). x =1

             x=1:\(\frac{3}{4}\)

             x=      \(\frac{4}{3}\)

b)\(\frac{4}{7}\).x=\(\frac{9}{8}\)-0,125

\(\frac{4}{7}\).x=1,125-0,125

\(\frac{4}{7}\).x=1

         x=1:\(\frac{4}{7}\)

         x=\(\frac{7}{4}\)

CHÚC BẠN HỌC GIỎI

K MÌNH NHÉ

x = 7/4

Nho tk cho minh nhe

= 2x² -7x -3x +3x -15 =0

=(2x² -10x) +(3x-15)  =0

2x(x-5) +3(x -5)    =0

(x-5)(2x+3) =0

x=5

x= -3/2

giữ lời hứa nhe bạn

2x² -7x-15 =0

2x² -7x -3x +3x -15 =0

(2x² -10x) +(3x-15)  =0

2x(x-5) +3(x -5)    =0

(x-5)(2x+3) =0

\(\Rightarrow\)(x-5)=0 hoac (2x+3)=0

          x =5              x = -3/2

I don't know

\(\frac{7x-8}{2x-3}\)đạt GTLN khi 2x - 3 = 1 => x = 2 và GTLN = 6 

Lời giải:

$3-2x-\frac{1}{3}=7x-\frac{1}{4}$

$3-\frac{1}{3}+\frac{1}{4}=2x+7x$

$\frac{35}{12}=9x$

$x=\frac{35}{108}$

\(3-2x-\dfrac{1}{3}=7x-\dfrac{1}{4}\)

\(\Leftrightarrow3-\dfrac{1}{3}+\dfrac{1}{4}=7x-2x\)

\(\Leftrightarrow5x=\dfrac{35}{12}\)

\(\Leftrightarrow x=\dfrac{7}{12}\)

( 3x-1) ( x²+ 9) = (3x-1) (7x-10)

⇒( 3x-1) ( x²+ 9) - (3x-1) (7x-10) = 0

⇒( 3x-1) (( x²+ 9)-(7x-10)) = 0

⇒( 3x-1)(x²+9-7x+10)=0

⇒( 3x-1)(x²-7x+19)=0

⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)

3x-1=0

⇒x=\(\dfrac{1}{3}\)

x²-7x+19=0

⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)

⇒ x vô nghiệm

Vậy x= \(\dfrac{1}{3}\)

\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Ta có: \(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+9-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+19\right)=0\)

\(\Leftrightarrow3x-1=0\)

hay \(x=\dfrac{1}{3}\)

\(6x^2-7x+1=6x^2-6x-x+1=6x\left(x-1\right)-\left(x-1\right)\\ =\left(6x-1\right)\left(x-1\right)\)

Lần sau gặp bài dạng này bn có cách làm như sau:

Dạng bài là \(ax^2+bx+c\) ta sẽ tách b sao cho b=a.c và \(b_1+b_2=b\)

Chúc bạn học tốt!!!

6x^2-7x+1=6x^2-6x-x+1=6x(x-1)-(x-1)=(x-1)(6x-1)

Ta có: 6x² - 7x + 1 = 6x² - 6x - x +1

= (6x² - 6x) - (x -1)

= 6x(x - 1) - (x - 1)

= (x - 1)(6x - 1)

\(a,=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x=20\)

a. \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\\ =\left(3x^2+12x\right)-\left(7x-20\right)+\left(2x^3-3x^2\right)-\left(2x^3+5x\right)\\ =3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\\ =\left(2x^3-2x^3\right)+\left(3x^2-3x^2\right)+\left(12x-7x-5x\right)+20\\ =0+0+0+20\\ =20\)