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LH
0
31 tháng 5 2017
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15+x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x-x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(A=\left|3x-2\right|\)
Dấu = xảy ra khi \(3x-2=0\)
\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
Amin = 1 <=> x = 2/3
c) \(B=\left|2x+3\right|\le5\)
Dấu = xảy ra <=> 2x + 3 = 0 => 2x = -3 => x = -3/2
Bmax = 5 <=> x = -3/2
3 tháng 6 2017
Bạn ơi hai câu b, c họ bảo tìm x chứ có phải tìm GTLN, NN đâu
FA
2
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)\\
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
PN
0
25 tháng 1 2020
a) Ta có: |4x+3|-x=15
⇒|4x+3|=15+x
\(\Rightarrow\left\{{}\begin{matrix}\left(4x+3\right)^2=\left(15+x\right)^2\\15+x\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}16x^2+24x+9=225+30x+x^2\\x\ge-15\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}15x^2-6x-216=0\\x\ge-15\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,6\end{matrix}\right.\)
Vậy: x∈{-3,6;4}
b) Ta có: |3x-2|-x>1
⇒|3x-2|>1+x
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>\frac{3}{2}\\x< \frac{1}{4}\end{matrix}\right.\)
Vậy: \(\frac{1}{4}< x< \frac{3}{2}\)
c) Ta có: \(\left|2x+3\right|\le5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\)
Vậy: \(-4\le x\le1\)
VM
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12:3\\x=\left(-18\right):5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\frac{3}{2}\\x< \frac{1}{4}\end{matrix}\right.\Rightarrow\frac{1}{4}< x< \frac{3}{2}.\)
Vậy \(\frac{1}{4}< x< \frac{3}{2}\) thì \(\left|3x-2\right|-x>1.\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\le2:2\\x\ge\left(-8\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\) thì \(\left|2x+3\right|\le5.\)
Chúc bạn học tốt!
3 tháng 11 2017
a) Ta có: |4x - 1| - x = 15
- Nếu \(4x-1\ge0\) \(\Rightarrow x\ge\frac{1}{4}\)
=> 4x - 1 - x = 15
=> 3x = 15 + 1
=> 3x = 16
=> x = \(\frac{16}{3}\) (thỏa mãn điều kiện)
- Nếu \(4x-1< 0\Rightarrow x< \frac{1}{4}\)
=> 1 - 4x - x = 15
=> -5x = 14
=> x = \(\frac{-14}{5}\) (thỏa mãn điều kiện)
Vậy x = \(\frac{16}{3}\) hoặc x = \(\frac{-14}{5}\)
3 tháng 11 2017
Câu b hình như là đề sai rùi bạn ơi.
c) Ta có: 2x = 3y
\(\Rightarrow\) \(\frac{x}{3}=\frac{y}{2}\) \(\Rightarrow\) \(\frac{x}{21}=\frac{y}{14}\) (1)
5y = 7z
\(\Rightarrow\) \(\frac{y}{7}=\frac{z}{5}\) \(\Rightarrow\) \(\frac{y}{14}=\frac{z}{10}\) (2)
Từ (1) và (2) suy ra:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
\(\Rightarrow\) \(\frac{x}{21}=2\) \(\Rightarrow\) \(x=21.2=42\)
\(\Rightarrow\) \(\frac{y}{14}=2\) \(\Rightarrow\) \(y=14.2=28\)
\(\Rightarrow\)\(\frac{z}{10}=2\) \(\Rightarrow\) \(z=10.2=20\)
Vậy x = 42; y = 28; z = 20