Gì vậy bạn

kết quả thi tiếng việt vòng 1 bn nhé

thấy đâu brooooooo

Ta có :

\(B=\left|5x-200\right|+\left|5x+1\right|=\left|200-5x\right|+\left|5x+1\right|\)

Với \(\forall x\) ta có :

\(B=\left|200-5x\right|+\left|5x+1\right|\)

\(\Leftrightarrow B\ge\left|200-5x+5x+1\right|\)

\(\Leftrightarrow B\ge\left|199\right|\)

\(\Leftrightarrow B\ge199\)

Dấu "=" xảy ra khi và chỉ khi :

\(\left(200-5x\right)\left(5x+1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}200-5x\ge0\\5x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}200-5x\le0\\5x+1\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le40\\x\ge-\dfrac{1}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge40\\x\le-\dfrac{1}{5}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}40\ge x\ge-\dfrac{1}{5}\\x\in\varnothing\end{matrix}\right.\)

Vậy ...

Bài 11:

+)\(A=10x^2+20xy+10y-90=10.\left(x^2+2xy+y\right)-10.9\)

\(=10.\left(x+y\right)^2-10.9=10.\left[\left(x+y\right)^2-9\right]\)

+)\(B=x^3y-3x^2y-4xy+12y=\left(x^3y-3x^2y\right)-\left(4xy-12y\right)\)

\(=x^2y.\left(x-3\right)-4y.\left(x-3\right)=\left(x^2y-4y\right)\left(x-3\right)\)

\(=y.\left(x^2-4\right)\left(x-3\right)=y.\left(x+2\right)\left(x-2\right)\left(x-3\right)\)

+) \(C=125x^3-10x^2+2x-1=\left(125x^3-1\right)-\left(10x^2-2x\right)\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)

\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)

Bài 12:

1) \(x^3-7x^2-9x+63=0\Rightarrow\left(x^3-7x^2\right)-\left(9x-63\right)=0\)

\(\Rightarrow x^2.\left(x-7\right)-9.\left(x-7\right)=0\Rightarrow\left(x-3\right)\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=7\end{matrix}\right.\)

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

\(\Rightarrow\left(x-1\right)^3+2x\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-1+2x\right)=0\)

\(\Rightarrow\left(x-1\right)\left(3x-1\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

anh nghĩ sao khi tag e vào bài này! Lớp 7 chưa học sao làm lớp 8 giờ! Để e nhờ ng` khác.

Nguyễn Huy Tú Tuấn Anh Phan Nguyễn văn tài Nguyễn Thị Huyền Trang

a) \(\left|\dfrac{1}{2}x\right|=3-2x\)

+) TH1: \(\dfrac{1}{2}x\ge0\Rightarrow x\ge0\)

Khi đó: \(\dfrac{1}{2}x=3-2x\)

\(\Rightarrow x=2\left(3-2x\right)\)

\(\Rightarrow x=6-4x\)

\(\Rightarrow x+4x=6\)

\(\Rightarrow5x=6\Rightarrow x=\dfrac{6}{5}\) (thỏa mãn)

+) TH2: \(-\dfrac{1}{2}x< 0\Rightarrow x>0\)

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

\(\Rightarrow-x=6-4x\)

\(\Rightarrow-x+4x=6\)

\(\Rightarrow3x=6\Rightarrow x=2\) (t/m)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\).

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

+) TH1: \(x-1\ge0\Rightarrow x\ge1\)

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

\(\Rightarrow x-3x=1+2\)

\(\Rightarrow-2x=3\)

\(\Rightarrow x=\dfrac{-3}{2}\) (loại)

+) TH2: \(x-1< 0\Rightarrow x< 1\)

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

\(\Rightarrow-x-3x=-1+2\)

\(\Rightarrow-4x=1\)

\(\Rightarrow x=\dfrac{-1}{4}\) (t/m)

Vậy \(x=-\dfrac{1}{4}.\)

c) Tương tự.

a) Vì \(\left|\dfrac{1}{2}x\right|\ge0\)

\(\Leftrightarrow3-2x\ge0\)

Mà \(\left|\dfrac{1}{2}x\right|=3-2x\)

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

\(\Leftrightarrow0,5x=3-2x\)

\(\Leftrightarrow0,5x+2x=3\)

\(\Leftrightarrow2,5x=3\)

\(\Leftrightarrow x=\dfrac{3}{2,5}\)

Vậy ................

b) Vì \(\left|x-1\right|\ge0\)

Mà \(\left|x-1\right|=3x+2\)

\(\Leftrightarrow3x+2\ge0\)

\(\Leftrightarrow x-1=3x+2\)

\(\Leftrightarrow x-3x=2+1\)

\(\Leftrightarrow-2x=3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)

Vậy ....................

c) Vì \(\left|5x\right|\ge0\)

Mà \(\left|5x\right|=x+12\)

\(\Leftrightarrow x+12\ge0\)

\(\Leftrightarrow5x=x+12\)

\(\Leftrightarrow5x-x=12\)

\(\Leftrightarrow4x=12\)

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

Vậy ............

Bà tag tui ko có trúng :v

a) \(\left|\dfrac{1}{2}x\right|=3-2x\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\left(x\ge0\right)\\\dfrac{1}{2}x=2x-3\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=2\end{matrix}\right.\)

\(\Rightarrow x=1,2\)

các bài khác tương tự nhé !!

Nguyễn Thanh Hằng banj owi

A

A

câu hỏi bị thiếu ??

Tỉ số là 3/5 hoặc 5/3

Tổng số phần bằng nhau là 

3+5=8 (phần)

Số tiền của Hằng là 

(60800÷8)×3=22800 (đồng)

Số tiền của Thanh là 

60800-22800=38000 (đồng)

Đ/s : ....

       : ....

Ta có: \(\left|3x-5\right|+\left|3x+1\right|=\left|5-3x\right|+\left|3x+1\right|\ge\left|5-3x+3x+1\right|=6\)

Dấu "=" xảy ra \(\Leftrightarrow\left(5-3x\right)\left(3x+1\right)\ge0\Leftrightarrow-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)

Vậy \(-\dfrac{1}{3}\le x\le\dfrac{5}{3}\)