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Bất phương trình bậc nhất 2 ẩn :
\(2x+3y>0\Rightarrow Câu\) \(C\)
\(x-2y\le1\Rightarrow Câu\) \(f\)
\(4\left(x-1\right)+5\left(y-3\right)>2x-9\)
\(\Leftrightarrow4x-4+5y-15-2x+9>0\)
\(\Leftrightarrow2x+5y-10>0\) \(\Rightarrow Câu\) \(i\)
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
Song song với d nên có a = 3
=> Ý B hoặc C
Thay x = 1; y = -2 vào câu B thấy thỏa mãn
Vậy Chọn B
Bài 1
d, \(x^2+2xy+y^2-2x-2y+1\)
\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)
\(\Rightarrow\left(x+y-1\right)^2\)
Bài 2:
a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\)
b,\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
c, \(4x^2-9=0\)
\(\Leftrightarrow4x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)
d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)
\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)
\(\Leftrightarrow7x^2-16x+9=0\)
\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)
\(\Leftrightarrow x=\frac{16\pm2}{14}\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)
1.a)\(3x-3y+x^2-2xy+y^2\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
d)\(x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y+1\right)^2\)
2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)
\(\Leftrightarrow-5x-9=0\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)
b)\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)
c)\(4x^2-9=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)
d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)
\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)
3.Ta có:
Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)
\(\Rightarrow m+21=0\)
\(\Rightarrow m=-21\)
Vậy...!
a)
\(A=3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)\(2A=\left[\left(x-y\right)-\left(x+y\right)\right]^2+5\left(x-y\right)^2-5\left(x+y\right)^2\)
\(2A=4y^2+5\left[\left(x-y\right)-\left(x+y\right)\right]\left[\left(x-y\right)+\left(x+y\right)\right]\)\(2A=4y^2+5\left[-2y\right]\left[2x\right]=4y^2-20xy=4y\left(y-5x\right)\\ \)\(A=2y\left(y-5x\right)\)
Bài 1:
a) \(5x-15y=5\left(x-3y\right)\)
b) \(\dfrac{3}{5}x^2+5x^4-x^2y=x^2\left(\dfrac{3}{5}+5x^2-y\right)\)
c) \(14x^2y^2-21xy^2+28x^2y=7xy\left(2xy-3y+4x\right)\)
d) \(\dfrac{2}{7}x\left(3y-1\right)-\dfrac{2}{7}y\left(3y-1\right)=\dfrac{2}{7}\left(3y-1\right)\left(x-y\right)\)
e) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
f) \(\left(x+y\right)^2-4x^2=\left(-x+y\right)\left(3x+y\right)\)
g) \(27x^3+\dfrac{1}{8}=\left(3x+\dfrac{1}{2}\right)\left(6x^2+1,5x+\dfrac{1}{4}\right)\)
h) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y\right)\)
Bài 2:
a) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\Rightarrow x=-1\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)
b) \(x\left(3x-2\right)-5\left(2-3x\right)=0\)
\(\Rightarrow x\left(3x-2\right)+5\left(3x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\Rightarrow x=\dfrac{2}{3}\\x+5=0\Rightarrow x=-5\end{matrix}\right.\)
c) \(\dfrac{4}{9}-25x^2=0\)
\(\Rightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\Rightarrow x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0\Rightarrow x=\dfrac{-2}{15}\end{matrix}\right.\)
d) Có tới 2 dấu "=".
bài 1 dễ mk ko lm nữa nhé
bafi2:
a,x(x+1)(x+2)=0
x=0 ; x=-1 ; x=-2
b,x(3x-2)+5(3x-2)=0
(x+5)(3x-2)=0
x=-5 ; x=2/3
c,
(2/3)2- (5x)2=0
(2/3-5x)(2/3+5x)=0
x=+-2/15
d, X2-2*1/2x+(1/2)2=0
(X-1/2)22=0
X=1/2