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\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a)x5+2x4+3x3+3x2+2x+1=0
<=> x5+x4+x4+x3+2x3+2x2+x2+x+x+1=0
<=>x4(x+1)+x3(x+1)+2x2(x+1)+x(x+1)+(x+1)=0
<=>(x+1)(x4+x3+2x2+x+1)=0
<=>x2(x+1)(x2+x+2+\(\dfrac{1}{x^2}\))=0
<=>x2(x+1)[(x+\(\dfrac{1}{2}\))2+\(\dfrac{7}{4}+\dfrac{1}{x^2}\)]=0
Vì [(x+\(\dfrac{1}{2}\))2\(+\dfrac{7}{4}+\dfrac{1}{x^2}\)]>0 với mọi x thuộc R
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy S={0;-1}
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
Giải phương trình
a.|x+4|−2|2x+3|=3−3x (1)
Lập bảng xét dấu
x -4 \(\dfrac{-3}{2}\)
x+4 - 0 + +
2x+3 - - 0 +
- Với \(x\le-4\) thì (1)
<=> -(x+4)+2(2x+3)=3-3x
<=> -x-4+4x+6=3-3x
<=> -x+4x+3x=4-6+3
<=> 6x=1
<=> x=\(\dfrac{1}{6}\) (L)
- Với \(-4\le x\le\dfrac{-3}{2}\) thì (1)
<=> (x+4)+2(2x+3)=3-3x
<=> x+4+4x+6=3-3x
<=> x+4x+3x=-4-6+3
<=> 8x=-7
<=> x=\(\dfrac{-7}{8}\) (L)
- Với \(x\ge\dfrac{-3}{2}\) thì (1)
<=> x+4-2(2x+3)=3-3x
<=> x+4-4x-6=3-3x
<=> x-4x+3x=-4+6+3
<=> 0x=5
<=> x (vô nghiệm) (L)
Vậy \(S=\varnothing\)
b.3|x−1|+|x−3|=x+5 (2)
Lập bảng xét dấu
x 1 3
x+1 - 0 + +
x-3 - - 0 +
+ Với \(x\le1\) thì (2)
<=> -3(x-1)-(x-3)=x+5
<=> -3x+3-x+3=x+5
<=> -3x-x-x=-3-3+5
<=> -5x=-1
<=> x= \(\dfrac{1}{5}\) (N)
+ Với \(1\le x\le3\) thì (2)
<=> 3(x-1)-(x-3)=x+5
<=> 3x-3-x+3=x+5
<=> 3x-x-x=3-3+5
<=> x=5(L)
+ Với \(x\ge3\) thì (2)
<=> 3(x-1)+(x-3)=x+5
<=> 3x-3+x-3=x+5
<=> 3x+x-x=3+3+5
<=> 3x=11
<=> x=\(\dfrac{11}{3}\) (N)
Vậy \(S=\left\{\dfrac{1}{5};\dfrac{11}{3}\right\}\)
Giải:
a) \(\left|x+4\right|-2\left|2x+3\right|=3-3x\)
\(\Leftrightarrow x+4-2\left(2x+3\right)=3-3x\)
\(\Leftrightarrow x+4-4x-6=3-3x\)
\(\Leftrightarrow x-4x+3x=3+6-4\)
\(\Leftrightarrow0x=5\)
Vậy phương trình vô nghiệm
b) \(3\left|x-1\right|+\left|x-3\right|=x+5\)
\(\Leftrightarrow3\left(x-1\right)+x-3=x+5\)
\(\Leftrightarrow3x-3+x-3=x+5\)
\(\Leftrightarrow3x+x-x=5+3+3\)
\(\Leftrightarrow3x=11\)
\(\Leftrightarrow x=\dfrac{11}{3}\)
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