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\(xy+2x+2y=-16\)
\(\Rightarrow xy+2x+2y+4=-12\)
\(\Rightarrow x\left(y+2\right)+2\left(y+2\right)=-12\)
\(\Rightarrow\left(x+2\right)\left(y+2\right)=-12\)
Xét ước 12 là xong mấy câu kia tương tự
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: \(-4\le x\le1\)
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!
Câu 1:
a)
Ta có: \(P\left(x\right)=5x^4+3x^3-6x+x^2-5x^4+2x+8\)
\(=3x^3+x^2-4x+8\)
Ta có: \(Q\left(x\right)=2x^2-3x^3+12-3x^2+6x^3-4\)
\(=-3x^3-x^2+8\)
b) Ta có: P(x)+Q(x)
\(=3x^3+x^2-4x+8-3x^3-x^2+8\)
\(=-4x+16\)
Ta có: H(x)+P(x)=Q(x)
⇔H(x)=Q(x)-P(x)
\(\Leftrightarrow H\left(x\right)=-3x^3-x^2+8-\left(3x^3+x^2-4x+8\right)\)
\(\Leftrightarrow H\left(x\right)=-3x^3-x^2+8-3x^3-x^2+4x-8\)
\(\Leftrightarrow H\left(x\right)=-6x^3-2x^2+4x\)
c) Đặt H(x)=0
\(\Leftrightarrow-6x^3-2x^2+4x=0\)
\(\Leftrightarrow x\left(-6x^2-2x+4\right)=0\)
\(\Leftrightarrow x\left(-6x^2-6x+4x+4\right)=0\)
\(\Leftrightarrow x\left[-6x\left(x+1\right)+4\left(x+1\right)\right]=0\)
\(\Leftrightarrow x\cdot\left(x+1\right)\cdot\left(-6x+4\right)=0\)
\(\Leftrightarrow-2\cdot\left(3x-2\right)\cdot x\cdot\left(x+1\right)=0\)
mà \(-2\ne0\)
nên \(\left[{}\begin{matrix}3x-2=0\\x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=0\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=0\\x=-1\end{matrix}\right.\)
Vậy: Nghiệm của đa thức H(x) lần lượt là 0;-1;\(\frac{2}{3}\)
Câu 2: Sửa đề: \(C=4x^2+7xy-3y^2\)
Ta có: A+B+C
=\(7x^2-12xy+9y^2+5-10x^2+7xy-5y^2+4x^2+7xy-3y^2\)
\(=x^2+2xy+y^2+5\)
\(=\left(x+y\right)^2+5>0\forall x,y\)(đpcm)
Bạn ơi bên trên mik viết nhầm câu 2 phần C = 4x\(^2\) + 7xy + 5y\(^2\)
b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)
\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)
\(\left|4-7x\right|=\dfrac{49}{30}\) (*)
+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)
PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)
\(-7x=-\dfrac{71}{30}\)
x = \(\dfrac{71}{210}\) (t/m)
+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)
Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)
x = \(\dfrac{169}{210}\) (t/m)
Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)
Bài giải
\(a,\text{ }\left|3x-2\right|-x>1\)
\(\left|3x-2\right|>x+1\)
TH1 : 3x - 2 < 0 => 3x < 3 => x < 1 thì :
\(3x-2>-x-1\)
\(3x+x>2-1\)
\(4x>1\)
\(x>\frac{1}{4}\)
=> \(\frac{1}{4}< x< 1\)
TH2 : 3x - 2 \(\ge\)0 => 3x \(\ge\)2 => x \(\ge\) \(\frac{2}{3}\) thì :
\(3x-2>x+1\)
\(3x-x>1+2\)
\(2x>3\)
\(x>\frac{3}{2}\)
Vậy \(\frac{1}{4}< x< 1\) hoặc \(x>\frac{3}{2}\)
\(A=\frac{x\left|x-2\right|}{x^2+8x-20}=\frac{x\left|x-2\right|}{x^2-2x+10x-20}=\frac{x\left|x-2\right|}{x\left(x-2\right)+10\left(x-2\right)}=\frac{x\left|x-2\right|}{\left(x+10\right)\left(x-2\right)}\)
Xét \(x-2\ge0\Leftrightarrow x\ge2\) ta có :
\(A=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}=\frac{x}{x+10}\)
Xét \(x-2< 0\Leftrightarrow x< 2\) ta có :
\(A=\frac{x\left(2-x\right)}{\left(x+10\right)\left(x-2\right)}=\frac{-x}{x+10}\)