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a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
\(a,\left(x-1\right)^2+x^2\le\left(x+1\right)^2+\left(x+2\right)^2\\ \Leftrightarrow x^2-2x+1+x^2\le x^2+2x+1+x^2+4x+4\\ \Leftrightarrow2x^2-2x+1\le2x^2+6x+5\\ \Leftrightarrow-8x-6\le0\\ \Leftrightarrow x\ge\dfrac{3}{4}\)
\(b,\left(x^2+1\right)\left(x-6\right)\le\left(x-2\right)^3\\ \Leftrightarrow x^3+x-6x^2-6\le x^3-6x^2+12x-8\\ \Leftrightarrow11x-2\ge0\\ \Leftrightarrow x\ge\dfrac{2}{11}\)
a: \(\Leftrightarrow x^2-2x+1+x^2< =x^2+2x+1+x^2+4x+4\)
=>-2x+1<=6x+5
=>-7x<=4
hay x>=-4/7
b: \(\Leftrightarrow x^3-6x^2+x-6-x^3+6x^2-12x+8< =0\)
=>-11x+2<=0
=>-11x<=-2
hay x>=2/11
a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)
\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)
=>3x-5<=30x-100
=>30x-100>3x-5
=>27x>95
hay x>95/27
b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)
=>26x-8<-11x
=>37x<8
hay x<8/37
a: \(\Leftrightarrow x^2+4x+4+3x^2+6x+3>=4x^2-4\)
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: \(\Leftrightarrow6\left(x-1\right)-4\left(x-2\right)\le12x-3\left(x-3\right)\)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1
a: ⇔x2+4x+4+3x2+6x+3>=4x2−4⇔x2+4x+4+3x2+6x+3>=4x2−4
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: ⇔6(x−1)−4(x−2)≤12x−3(x−3)⇔6(x−1)−4(x−2)≤12x−3(x−3)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1
a, Xét 2 trường hợp: x+1/9<0
2x-5<0
Tự làm nốt nhé, chuyển vế mà k bít làm thì mình bó tay.
b, Tương tự câu a, nhưng chọn 1 cái âm và 2 cái còn lại dương
VD: Xét 4x-1 âm, còn lại dương
TỰ LÀM NỐT ĐI, CHUYỂN VẾ NHÉ. BẤM NÚT ĐÚNG Ở PHÍA DƯỚI ĐẤY
a: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2\)
b: \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3+2x^2-x-2-x^3+8\)
\(=2x^2-x+6\)
a) \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=\left(4x\right)^2\)
\(=16x^2\)
b) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)\)
\(=x^3+2x^2-x-2-x^3+8\)
\(=2x^2-x+6\)
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
a: (2x+3)(x+1)<0
=>2x+3 và x+1 khác dấu
=>x>-1 hoặc x<-3/2
b: (4-x)(x+2)>0
=>(x-4)(x+2)<0
=>-2<x<4
a: (2x+3)(x+1)<0
=>2x+3 và x+1 khác dấu
=>x>-1 hoặc x<-3/2
b: (4-x)(x+2)>0
=>(x-4)(x+2)<0
=>-2<x<4