\(\dfrac{x}{6}=\dfrac{7}{4}\Rightarrow x=\dfrac{6\cdot7}{4}=\dfrac{21}{2}\\ \dfrac{3}{x}=\dfrac{21}{17}\Rightarrow x=\dfrac{3\cdot17}{21}=\dfrac{17}{7}\)

\(12\left(x-1\right)=0\\ =>x-1=0:12\\ =>x-1=0\\ =>x=0+1\\ =>x=1\)

\(12.\left(x-1\right)=0\)

\(x-1=0:12\)

\(x-1=0\)

\(x=0+1\)

\(x=1\)

12 . (x-1) = 0

(x-1) = 0 : 12

(x-1) = 0

x = 0 + 1

x = 1

Vậy x = 1

Vì | x+5 | >=0 với mọi x

    | y - 4 | >=0 với mọi y

=> |x +5| +|y-4|>=0

Mà |x+5|+|y-4|<=0 

=> \(\hept{\begin{cases}x+5=0\\y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=4\end{cases}}\)

vậy ...........

hok tốt

Ta có: \(\hept{\begin{cases}\left|x+5\right|\ge0\forall x\\\left|y-4\right|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|x+5\right|+\left|y-4\right|\ge0\)

 \(\Rightarrow\left|x+5\right|+\left|y+4\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left|x+5\right|=0\\\left|y-4\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+5=0\\y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-5\\y=4\end{cases}}}\)

vậy x =-5; y = 4

hok tốt!!

|x+5|+|y-4| < 0

Vì |x+5| > 0 (với mọi x)

     |y-4| > 0 (với mọi y)

=> |x+5|+|y-4| > 0

Mà |x+5|+|y-4| < 0

=> x+5=0 hoặc y-4=0

=      x=-5 hoặc     y=-4

Vậy x=5 hoặc y=-4

<=> /x-1/=x-1

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

vậy x>1 thì thõa mãn đề bài

\(\left(2.x+\frac{1}{3}\right)^2=\frac{16}{25}\)

\(\Leftrightarrow2.x+\frac{1}{3}=\pm\sqrt{\frac{16}{25}}\)

\(\Leftrightarrow2.x+\frac{1}{3}=\pm\frac{4}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}2.x+\frac{1}{3}=\frac{4}{5}\\2.x+\frac{1}{3}=-\frac{4}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2.x=\frac{7}{15}\\2.x=-\frac{17}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{30}\\x=-\frac{17}{30}\end{cases}}\)

\(\left(2.x+\frac{1}{3}\right)^2=\frac{16}{25}\)

\(\left(2.x+\frac{1}{3}\right)^2=\left(\frac{4}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2.x+\frac{1}{3}=\frac{4}{5}\\2.x+\frac{1}{3}=\frac{-4}{5}\end{cases}\Rightarrow\orbr{\begin{cases}2.x=\frac{4}{5}-\frac{1}{3}\\2.x=\frac{-4}{5}-\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}2.x=\frac{12}{15}-\frac{5}{15}\\2.x=\frac{-12}{15}-\frac{5}{15}\end{cases}\Rightarrow}\orbr{\begin{cases}2.x=\frac{7}{15}\\2.x=\frac{-17}{15}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}:2\\x=\frac{-17}{15}:2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}.\frac{1}{2}\\x=\frac{-17}{15}.\frac{1}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{30}\\x=\frac{-17}{30}\end{cases}}}\)

Vậy \(x=\frac{7}{30}\)hoặc \(x=\frac{-17}{30}\)

a) \(x^3+2x^2-4x+1\)

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

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

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

c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp

Tham khảo

b) Phân tích đa thức A=(x^2+x+4)^2+8x(x^2+x+4)+15x^2 thành nhân tử - Nguyễn Hồng Tiến

Tham khảo