Tìm x biết:
a) \(x+y=4\) và \(\left|2x+1\right|+\left|y-x\right|=5\)
b) \(x-y=3\) và \(\left|x\right|+\left|y\right|=3\)
Tìm x biết:
a) \(x+y=4\) và \(\left|2x+1\right|+\left|y-x\right|=5\)
b) \(x-y=3\) và \(\left|x\right|+\left|y\right|=3\)
Tìm các số thực x và y, biết :
a) \(\left(3x-2\right)+\left(2y+1\right)i=\left(x+1\right)-\left(y-5\right)i\)
b) \(\left(1-2x\right)-i\sqrt{3}=\sqrt{5}+\left(1-3y\right)i\)
c) \(\left(2x+y\right)+\left(2y-x\right)i=\left(x-2y+3\right)+\left(y+2x+1\right)i\)
Từ định nghĩa bằng nhau của hai số phức, ta có:
a) ⇔ ;
b) ⇔ ;
c) ⇔ ⇔ .
1)tìm các số nguyên x và y thỏa mãn:\(y^2=x^2+x+1\)
2)cho các số thực x và y thỏa mãn \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)=a
tìm giá trị biểu thức \(4\left(x^7+y^7\right)+2\left(x^5+y^5\right)+11\left(x^3+y^3\right)+2016\)
3)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0
cmr \(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(=\frac{1}{x^3y^3}\)
4)cho a,b,c là các số dương.cmr\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
Tìm x:
a) 2x(x-5)-x(2x+3)=26
b) \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)
c) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)
c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow x=-\frac{3}{5}\)
Thực hiện phép tính
a,\(\left(x-y\right)\left(y^2+y+1\right)+\left(\dfrac{1}{3}x^2y-y\right)\left(2x+y^2\right)\)
b,\(2x^2\left(x-2\right)+3x\left(x^2-x-2\right)-5\left(3-x^2\right)\)
c,\(\left(x-1\right)\left(x-3\right)-\left(4-x\right)\left(2x-1\right)-3x^3+2x-5\)
a: \(=xy^2+xy+x-y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3-2xy-y^3\)
\(=xy^2-xy+x-2y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3\)
b: \(=2x^3-4x^2+3x^3-3x^2-6x-15+5x^2\)
\(=5x^3-2x^2-6x-15\)
c: \(=x^2-4x+3+\left(x-4\right)\left(2x-1\right)-3x^3+2x-5\)
\(=-3x^3+x^2-2x-2+2x^2-x-8x+4\)
\(=-3x^3+3x^2-11x+2\)
Bài 1: Rút gọn các biểu thức sau:
a) \(3x^2\) - 2x( 5+ 1,5x) +10
b) 7x ( 4y- x) + 4y( y-7x) - 2( \(2y^2\) - 3,5x)
c) \(\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}.\left(-2x\right)\)
Bài 2: Tìm x, biết:
a) 3( 2x -1) - 5( x -3) + 6( 3x -4) = 24
b) \(2x^2+3\left(x^2-1\right)=5x\left(x+1\right)\)
c) \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=3\)
d) \(3x\left(x+1\right)-2x\left(x+2\right)=-1-x\)
Bài 3: Tính giá trị của các biểu thức sau:
a)\(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002\) Với \(x=1;y=-1\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)-\dfrac{11}{20}\) Với \(x=-0,6;y=-0,75\)
Bài 4: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị biến:
a) \(2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)
b) \(z\left(y-x\right)+y\left(z-x\right)+x\left(y+z\right)-2yz+100\)
c) \(2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)
Bài 5: Tính giá trị của biểu thức:
a) \(A=\left(x-3\right)\left(x-7\right)-\left(2x-5\right)\left(x-1\right)\) Với \(x=0;x=1;x=-1\)
b) \(B=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\) Với \(\left|x\right|=2\)
c) \(C=\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\) Với \(x=1;y=1;z=\left|1\right|\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Bài 2:
c) \(2x(5-3x)+2x(3x-5)-3(x-7)=3\)
\(\Leftrightarrow 2x(5-3x)-2x(5-3x)-3(x-7)=3\)
\(\Leftrightarrow -3(x-7)=3\)
\(\Leftrightarrow x-7=-1\Rightarrow x=6\)
d)
\(3x(x+1)-2x(x+2)=-1-x\)
\(\Leftrightarrow 3x^2+3x-(2x^2+4x)+x+1=0\)
\(\Leftrightarrow x^2+1=0\)
Vô lý vì \(x^2+1\geq 0+1=1>0\) với mọi $x$
Vậy không tồn tại $x$ thỏa mãn.
bài 1 tìm x biết
a)\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
b)\(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
tìm khoảng đồng biến nghịch biến
a) \(y=\left(x^2-1\right)^2\)
b) \(y=\left(3x+4\right)^3\)
c) \(y=\left(x+3\right)^2\left(x-1\right)\)
d) \(y=\left(2x+2\right)\left(x^3-1\right)\)
a: \(y=\left(x^2-1\right)^2\)
=>\(y'=2\left(x^2-1\right)'\left(x^2-1\right)\)
\(=4x\left(x^2-1\right)\)
Đặt y'>0
=>\(x\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>\(x>1\)
TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< 0\)
Đặt y'<0
=>\(x\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x^2< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\-1< x< 1\end{matrix}\right.\)
=>0<x<1
TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>x<-1
Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-1;0\right)\)
Hàm số nghịch biến trên các khoảng (0;1) và \(\left(-\infty;-1\right)\)
b: \(y=\left(3x+4\right)^3\)
=>\(y'=3\left(3x+4\right)'\left(3x+4\right)^2\)
\(\Leftrightarrow y'=9\left(3x+4\right)^2>=0\forall x\)
=>Hàm số luôn đồng biến trên R
c: \(y=\left(x+3\right)^2\left(x-1\right)\)
=>\(y=\left(x^2+6x+9\right)\left(x-1\right)\)
=>\(y'=\left(x^2+6x+9\right)'\left(x-1\right)+\left(x^2+6x+9\right)\left(x-1\right)'\)
=>\(y'=\left(2x+6\right)\left(x-1\right)+x^2+6x+9\)
=>\(y'=2x^2-2x+6x-6+x^2+6x+9\)
=>\(y'=3x^2-2x+3\)
\(\Leftrightarrow y'=3\left(x^2-\dfrac{2}{3}x+1\right)\)
=>\(y'=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{8}{9}\right)\)
=>\(y'=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{3}>=\dfrac{8}{3}>0\forall x\)
=>Hàm số luôn đồng biến trên R
d: \(y=\left(2x+2\right)\left(x^3-1\right)\)
=>\(y'=\left(2x+2\right)'\left(x^3-1\right)+\left(2x+2\right)\left(x^3-1\right)'\)
\(=2\left(x^3-1\right)+3x^2\left(2x+2\right)\)
\(=2x^3-2+6x^3+6x^2\)
\(=8x^3+6x^2-2\)
Đặt y'>0
=>\(8x^3+6x^2-2>0\)
=>\(x>0,46\)
Đặt y'<0
=>\(8x^3+6x^2-2< 0\)
=>\(x< 0,46\)
Vậy: Hàm số đồng biến trên khoảng tầm \(\left(0,46;+\infty\right)\)
Hàm số nghịch biến trên khoảng tầm \(\left(-\infty;0,46\right)\)