1) -x - 3 = 7

-x = 7 + 3

-x = 10

x = -10

2) x + 5 = -10

x = -10 - 5

x = -15

3) 2x - 7 = 713

2x = 713 + 7

2x = 720

x = 720 : 2

x = 360

4) -129 - (35 - x) = 55

35 - x = -129 - 55

35 - x = -184

x = 35 - (-184)

x = 219

5) 103 - x = 16 - (13 - 8)

103 - x = 16 - 5

103 - x = 11

x = 103 - 11

x = 192

HPT CÓ 2 NGHIỆM

6.

a, ĐK: \(x\ne2;x\ne3\)

\(P=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\)

\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\dfrac{2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}\)

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

b, \(P=\dfrac{x+1}{x-3}=-\dfrac{1}{2}\Leftrightarrow2x+2=-x+3\Leftrightarrow x=\dfrac{1}{3}\)

\(P< 1\Leftrightarrow\dfrac{x+1}{x-3}< 1\Leftrightarrow\dfrac{4}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

a/ \(=\dfrac{2x-9-\left(x+3\right)\left(x-3\right)+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\) ( x khác 2 ; x khác 3)

\(=\dfrac{2x-9-x^2+9+2x^2+x-4x-2}{\left(x-2\right)\left(x-3\right)}\)

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

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

b/ \(P=\dfrac{1}{2}\Leftrightarrow\dfrac{x+1}{x-3}=\dfrac{1}{2}\)(x khác 2 ; x khác 3)

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

\(\Leftrightarrow x=-5\)

\(P< 1\Leftrightarrow\dfrac{x+1}{x-3}-1< 0\)

\(\Leftrightarrow\dfrac{4}{x-3}< 0\)

\(\Leftrightarrow x-3>0\Leftrightarrow x>3\) (tmđk)

c/ \(x^2-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-2\end{matrix}\right.\)

Thay x = -2 vào P ta có

\(P=\dfrac{-2+1}{-2-3}=-\dfrac{1}{-5}=\dfrac{1}{5}\)

d/ \(P=\dfrac{x+1}{x-3}\in Z^+\)

\(\Leftrightarrow1+\dfrac{4}{x-3}\) nguyên dương

\(\Leftrightarrow\dfrac{4}{x-3}\) nguyên dương

\(\Leftrightarrow x-3\in\left\{1;4\right\}\)

\(\Leftrightarrow x\in\left\{4;7\right\}\)

KHĐK x khác 2 ; x khác 3 ta có

x ∈ {4;7}

3.

\(y=\dfrac{1-sin^24x}{5}=\dfrac{cos^24x}{5}\)

\(cos4x\in\left[-1;1\right]\Rightarrow cos^24x\in\left[0;1\right]\Rightarrow y\in\left[0;\dfrac{1}{5}\right]\Rightarrow\left\{{}\begin{matrix}y_{min}=0\\y_{max}=\dfrac{1}{5}\end{matrix}\right.\)

6.

\(y=sinx+cosx+2=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\)

\(sin\left(x+\dfrac{\pi}{4}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\in\left[-\sqrt{2}+2;\sqrt{2}+2\right]\)

\(\Rightarrow y_{min}=-\sqrt{2}+2\)

\(y_{max}=\sqrt{2}+2\)

1.

\(sin2x\in\left[-1;1\right]\Rightarrow y=3-2sin2x\in\left[1;5\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=1\\y_{max}=5\end{matrix}\right.\)

và chỉ khi????Xin lỗi,mình hơi ngu nên ko hiểu cụm từ này!!