(2x+1)^2+(2-x)(2x+1)<=0
giải BPT
giúp e với, e cần luôn ạ!
(2x+1)^2+(2-x)(2x+1)<=0
giải BPT
giúp e với, e cần luôn ạ!
\(4x^2+4x+1+4x+2-2x^2-x\le0\)
\(\Leftrightarrow2x^2+7x+3\le0\Leftrightarrow\left(2x+1\right)\left(x+3\right)\le0\)
TH1 : \(\left\{{}\begin{matrix}2x+1\ge0\\x+3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-3\end{matrix}\right.\)<=> -1/2 =< x =< -3
TH2 : \(\left\{{}\begin{matrix}2x+1\le0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x\ge-3\end{matrix}\right.\)( vô lí )
(x+2)(3-4x)>x^2+4x+4
Giúp em với ạ! E cần gấp
\(3x-4x^2+6-8x>x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2>0\Leftrightarrow\left(5x-1\right)\left(x+2\right)>0\)
TH1 : \(\left\{{}\begin{matrix}5x-1>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x>-2\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{5}\)
TH2 : \(\left\{{}\begin{matrix}5x-1< 0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x< -2\end{matrix}\right.\Leftrightarrow x< -2\)
pt tương đương:
\(\left(x+3\right)\left(x-2\right)-\left(2-x\right)\left(6-x\right)>0\)
\(\Leftrightarrow-9\left(x-2\right)>0\)
đổi dấu thì suy ra : x<2
kí hiệu khoảng: \(\left\{1;0;-\infty\right\}\)
\(\Leftrightarrow\dfrac{3\left(3x+5\right)-6}{6}\le\dfrac{2\left(x+2\right)+6x}{6}\)
\(\Leftrightarrow9x+15-6\le2x+4+6x\)
\(\Leftrightarrow9x+9\le8x+4\)
\(\Leftrightarrow x\le-5\)
Vì ax+b=0 nhận x=1 làm nghiệm nên a+b=0
a: a=3 thì b=-3
b: b=5 thì a=-5
Bài 4:
Thay x=1 vào pt, ta được:
\(\left(3-a\right)\cdot2-\left(1+2\right)^2=1\)
\(\Leftrightarrow\left(3-a\right)\cdot2=1+9=10\)
=>3-a=5
hay a=-2
bài 3: giải phương trình
a) \(\dfrac{5x-7
}{3}=\dfrac{5-3x}{2}\)
b) \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
c) \(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
d) \(4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\)
a: =>10x-14=15-9x
=>19x=29
hay x=29/19
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x+9=32x+60
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>101x=101
hay x=1
d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)
\(\Leftrightarrow6-18x+5x-6=0\)
=>-13x=0
hay x=0
\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)
\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)
\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)
\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)
Bài 1 : giải phương trình
a) (x-2)(x+2)-(2x+1)2=x(2-3x)
b) 2x(x+2)2-8x2=2(x-2)(x2+2x+4)
c) (x-2)3+(3x-1)(3x+1)=(x+1)3
d) 5(2x-3)-4(5x-7)=19-2(x+1)2
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
Bài 1 : giải phương trình
a) 7x-63=0
b)3x+15=0
c) 2x-5
d) -6x+16=0
a: =>7x=63
hay x=9
b: =>3x=-15
hay x=-5
d: =>-6x=-16
hay x=8/3
a) \(7x=63\Leftrightarrow x=9\)
b) \(3x=-15\Leftrightarrow x=-5\)
c) \(2x-5=0\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)
d) \(-6x=-16\Leftrightarrow x=\dfrac{8}{3}\)
help meeee
g: =>x-1-2x+1-9+x=0
=>-9=0(vô lý)
h: \(\Leftrightarrow x^2+x-12-6x+4-x^2+8x-16=0\)
=>3x-24=0
hay x=8
i: \(\Leftrightarrow x^3+6x^2+9x-3x-x^3-6x^2-12x-8-1=0\)
=>-6x-9=0
=>6x=-9
hay x=-3/2