Question

Tìm x biết:

a) |4x+3|-x=15

b) |3x-2|-x>1

c) |2x+3|5

0 câu trả lời

Toán lớp 7 Ôn tập toán 7

Answer 1

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: \(-4\le x\le1\)

Answer 2

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 \(\left|2x+3\right|\le5.\)

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