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\(\left(2-x\right)\left(2x-5\right)\)
Th1 : \(\hept{\begin{cases}2-x>0\\2x-5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x< \frac{5}{2}\end{cases}}}\)
Th2 : \(\hept{\begin{cases}2-x< 0\\2x-5>0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x>\frac{5}{2}\end{cases}}}\)
1) \(x^4-2x^2-144x+1295=0\)
\(\Rightarrow\)Cậu xem lại đề thử xem nhé !
2) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-1\right)-24=0\)
\(\Leftrightarrow x^4+2x^3-x^2-2x-24=0\)
\(\Leftrightarrow x^4+x^3+4x^2+x^3+x^2+4x-6x^2-6x-24=0\)
\(\Leftrightarrow x^2\left(x^2+x+4\right)+x\left(x^2+x+4\right)-6\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\)\(x+3=0\)
hoặc \(x-2=0\)
hoặc \(x^2+x+4=0\)
\(\Leftrightarrow\)\(x=-3\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)
3) \(x^4-2x^3+4x^2-3x-10=0\)
\(\Leftrightarrow x^4+x^3-3x^3-3x^2+7x^2+7x-10x-10=0\)
\(\Leftrightarrow x^3\left(x+1\right)-3x^2\left(x+1\right)+7x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-3x^2+7x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-2x^2-x^2+2x+5x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x-2\right)-x\left(x-2\right)+5\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)
hoặc \(x-2=0\)
hoặc \(x^2-x+5=0\)
\(\Leftrightarrow x=-1\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là :\(S=\left\{-1;2\right\}\)
a. 6x2 - (2x + 5)(3x - 2) = 7
<=> 6x2 - 6x2 + 4x - 15x + 10 = 7
<=> -11x = -3
<=> \(x=\dfrac{3}{11}\)
b. (5 - x)(25 + 5x + x2) + x(x2 - 7) = 25
<=> 125 - x3 + x3 - 7x = 25
<=> -7x = 25 - 125
<=> -7x = -100
<=> \(x=\dfrac{100}{7}\)
c. (7 - 2x)2 + (3 + 2x)(3 - 2x) = 30
<=> 49 - 28x + 4x2 + 9 - 4x2 = 30
<=> 4x2 - 4x2 - 28x = 30 - 49 - 9
<=> -28x = -28
<=> x = 1
\(5\left(x+3\right)-2x\left(x+3\right)=0\)
<=> \(\left(5-2x\right)\left(x+3\right)=0\)
<=> \(\hept{\begin{cases}5-2x=0\\x+3=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
\(4x\left(x-2018\right)-x+2018=0\)
<=> \(4x\left(x-2018\right)-\left(x-2018\right)=0\)
<=> \(\left(4x-1\right)\left(x-2018\right)=0\)
<=> \(\hept{\begin{cases}4x-1=0\\x-2018=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{4}\\x=2018\end{cases}}\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(x+1-1\right)=0\)
<=> \(\left(x+1\right).x=0\)
<=> \(\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
học tốt
a) \(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(5\left(x+3\right)+2x\left(x+3\right)=0\)
\(\left(x+3\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{-5}{2}\end{cases}}\)
b) \(4x\left(x-2018\right)-x+2018=0\)
\(4x\left(x-2018\right)-\left(x-2018\right)=0\)
\(\left(x-2018\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2018=0\\4x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{4}\end{cases}}\)
c) \(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+1-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
A= (x^2+2x)^2+9x^2+18x+20
=x^4+4x^4+4x^2+9x^2+18x+20
=5x^4+13x^2+18x+20
Cái bài này bạn yêu cầu không rõ nên mình chỉ giúp bạn được bấy nhiêu thôi. Nếu bạn yêu cầu rút gọn thì như trên còn yêu câu khác thì mình chưa chắc nên bạn phải ghi cụ thể nha.
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
3 - ( x2 + 2x )2 + 2x2 + 4x \(\ge\) 0 \(\Leftrightarrow\left(x^2+2x\right)^2+2\left(x^2+2x\right)-3\le0.\) Đặt t = x2 + 2x = (x + 1)2 - 1 , \(t\ge-1.\)
BPT trở thành : \(\hept{\begin{cases}t^2+2t-3\le0\\t=(x+1)^2-1\ge-1\end{cases}\Leftrightarrow\hept{\begin{cases}-3\le t\le1\\t\ge-1\end{cases}\Leftrightarrow}-1\le t\le1.}\)
Vậy ta có : \(-1\le x^2+2x\le1\Leftrightarrow x^2+2x-1\le0\Leftrightarrow-1-\sqrt{2}\le x\le-1+\sqrt{2}.\)