a

\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)

<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)

Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)

Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)

Vây x= 2 hoặc 3; 4

Đề tớ gõ sai, Sr các cậu...

Đề đúng là :

\(\frac{x-3}{90}+\frac{x-2}{91}+\frac{x-1}{92}=3\)

Giúp tớ nhen...Giải chi tiết giùm nha...Thank you !!!

\(\left(\frac{x-3}{90}-1\right)+\left(\frac{x-2}{91}-1\right)+\left(\frac{x-1}{90}-1\right)=0\)

\(\Leftrightarrow\frac{x-93}{90}+\frac{x-93}{91}+\frac{x-93}{92}=0\)

\(\Leftrightarrow\left(x-93\right)\left(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\right)=0\)

mà \(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\ne0\)

\(\Leftrightarrow x-93=0\Leftrightarrow x=93\)

Vậy x=93

a, \(x^2\) = \(x^3\) 

     \(x^3\) - \(x^2\) = 0

     \(x^2\)( \(x\) -1) = 0

      \(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy \(x\) \(\in\) { 0; 1}

e, 3^2\(x+1\) = 27

     \(3^{2x}\)⁺¹ = 3³

       2\(x\) + 1 = 3

        2\(x\)       = 2

         \(x\)         = 1

g, 6² = 6^\(x-3\)  

    2 = \(x-3\)

      \(x\) = 3 + 2

       \(x\) = 5

\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)

\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)

f(x)=9x³-1/3x+3x²-3x+1/3x²-1/9x³-3x²-9x+27+3x

    = 9x³-1/9x³+3x²+1/3x²-3x²-1/3-3x-9x+3x+27

   = 80/9x³+1/3x²-28/3x+27

cho hỏi tập hợp các số tự nhiên ko vượt quá 50

Bài 1:

câu a: 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

        = \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

        = \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

         = \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

          = \(\dfrac{29}{6}\)

b, (15,25 + 3,75) \(\times\) 4 + ( 20,71 + 5,29)\(\times\) 5

     = 19 \(\times\) 4 + 26 \(\times\) 5

     = 76 + 130

     = 206

c, \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{3}\) - \(\dfrac{4}{5}\) \(\times\) \(\dfrac{1}{4}\)

= \(\dfrac{2}{5}\)    + \(\dfrac{4}{15}\) - \(\dfrac{1}{5}\)

= \(\dfrac{6}{15}\) + \(\dfrac{4}{15}\) - \(\dfrac{3}{15}\)

= \(\dfrac{7}{15}\) 

d, 1\(\dfrac{5}{7}\) + 7\(\dfrac{3}{6}\) + 2\(\dfrac{2}{7}\) - 4\(\dfrac{3}{6}\)

= (1 + 2 + \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\)) + ( 7 + \(\dfrac{3}{6}\) - 4 - \(\dfrac{3}{6}\))

=  3 + 1 + 3 

= 7

e, 8,4 \(\times\) \(x\) + 1,6 \(\times\) \(x\) = 10

   (8,4 + 1,6) \(\times\) \(x\)      = 10

     10 \(\times\) \(x\)                = 10

               \(x\)                = 1

\(3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x^2-3x\right)+\left(-2x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

xin lỗi toán lớp 8 thì mk chịu

\(3\left(6x^2-5x-6\right)-2\left(4x^2-13x-12\right)+36x-9x^2=0\)

\(10x^2-15x-18-8x^2+26x+24+36x-9x^2=0\)

\(-7x^2+47x+6=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-47+\sqrt{2377}}{-14}\\x=\frac{-47-\sqrt{2377}}{-14}\end{cases}}\)

a, \(\dfrac{x}{y}=\dfrac{17}{3}\&x+y=-60\)

Từ \(\dfrac{x}{y}=\dfrac{17}{3}\Rightarrow\dfrac{x}{17}=\dfrac{y}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{17}=\dfrac{y}{3}=\dfrac{x+y}{17+3}=\dfrac{-60}{20}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{17}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-51\\y=-9\end{matrix}\right.\)

b, \(\dfrac{x}{19}=\dfrac{y}{21}\&2x-y=34\)

\(\dfrac{x}{19}=\dfrac{y}{21}\Leftrightarrow\dfrac{2x}{2.19}=\dfrac{y}{21}\Leftrightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x}{38}=2\\\dfrac{y}{21}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=38\\y=42\end{matrix}\right.\)

c, \(\dfrac{x^2}{9}=\dfrac{y^2}{16}\&x^2+y^2=100\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{9}=4\\\dfrac{y^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\end{matrix}\right.\)

Các cặp (x;y) tương ứng là: \(\left(6;8\right)\&\left(-6;-8\right)\)

a) x/y=17/3=>x/17=y/3=x+y/17+3=60/20=3

Vậy: x=3.17=51; y=3.3=9

b)x/19=y/21=2x-y/19.2-21=34/17=2

Vậy: x=2.19=38; y=2.21=42

c)x^2/9=y^2/16=x^2+y^2/9+16=100/25=4

Vậy: x^2=36=>x=6; y^2=64=>y=8