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a, - Đặt \(x^2+x=a\) ta được phương trình :\(a^2+4a-12=0\)
=> \(a^2-2a+6a-12=0\)
=> \(a\left(a-2\right)+6\left(a-2\right)=0\)
=> \(\left(a+6\right)\left(a-2\right)=0\)
=> \(\left[{}\begin{matrix}a+6=0\\a-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=2\\a=-6\end{matrix}\right.\)
- Thay lại \(x^2+x=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+6=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{9}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\left(VL\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{9}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{9}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{\frac{9}{4}}-\frac{1}{2}=1\\x=-\sqrt{\frac{9}{4}}-\frac{1}{2}=-2\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(S=\left\{1,-2\right\}\)
b, Đặt \(x^2+2x+3=a\) -> làm tương tự câu a .
c, Ta có : \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
=> \(\left(x^2-4\right)\left(x^2-10\right)=72\)
- Đặt \(x^2-4=a\) và \(x^2-10=a-6\) ta được phương trình :
\(a\left(a-6\right)=72\)
=> \(a^2-6a-72=0\)
=> \(a^2+6a-12a-72=0\)
=> \(a\left(a+6\right)-12\left(a+6\right)=0\)
=> \(\left(a+6\right)\left(a-12\right)=0\)
=> \(\left[{}\begin{matrix}a+6=0\\a-12=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a=-6\\a=12\end{matrix}\right.\)
- Thay lại \(x^2-4=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2-4=-6\\x^2-4=12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=-2\left(VL\right)\\x^2=16\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{16}=4\\x=-\sqrt{16}=-4\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(S=\left\{4,-4\right\}\)
d, Ta có : \(x\left(x+1\right)\left(x^2+x+1\right)=42\)
=> \(\left(x^2+x\right)\left(x^2+x+1\right)=42\)
- Đặt \(x^2+x=a\) ta được phương trình : \(a\left(a+1\right)=42\)
=> \(a^2+a-42=0\)
=> \(a^2+7a-6a-42=0\)
=> \(a\left(a+7\right)-6\left(a+7\right)=0\)
=> \(\left(a-6\right)\left(a+7\right)=0\)
=> \(\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)
- Thay \(a=x^2+x\) vào phương trình ta được : \(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{25}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{27}{4}\left(VL\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{25}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{25}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{\frac{25}{4}}-\frac{1}{2}=2\\x=-\sqrt{\frac{25}{4}}-\frac{1}{2}=-3\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2;-3\right\}\)
pt <=> (x^4-5x^3)+(5x^3-25x^2)+(23x^2-115x)+(259x-1295) = 0
<=> (x-5).(x^3+5x^2+23x+259) = 0
<=> (x-5).[(x^3+7x^2)-(2x^2+14x)+(37x+259)] = 0
<=> (x-5).(x+7).(x^2-2x+37) = 0
<=> (x-5).(x+7) = 0 ( vì x^2-2x+37 > 0 )
<=> x-5=0 hoặc x+7=0
<=> x=5 hoặc x=-7
Vậy .........
Tk mk nha
câu a sai đề nha
Nếu câu a đề đúng thì phương trình vô nghiệm nha
Theo mình đây là đề đúng
\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)-24=0\)
Đặt a=\(\left(2x^2+3x-1\right)\)
Khi đó, phương trình trở thành
\(a^2-5a-24=0\)
\(\left(a-8\right)\left(a+3\right)=0\)
\(\left[{}\begin{matrix}a=8\\a=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-1=8\\2x^2+3x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-9=0\\2x^2+3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(2x-3\right)=0\\2\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{8}=0\left(vl\right)\end{matrix}\right.\)\(\left[{}\begin{matrix}x=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)
=>x^4+2x^2+1-4x^2-144x-1296=0
=>(x^2+1)^2-(2x+36)^2=0
=>(x^2+1-2x-36)(x^2+1+2x+36)=0
=>x^2-2x-35=0
=>(x-7)(x+5)=0
=>x=7 hoặc x=-5
a/ Đặt \(x^2+2x+1=\left(x+1\right)^2=t\ge0\)
\(\Rightarrow\left(t+2\right)t=3\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=1\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b/ \(\Leftrightarrow\left(x^2-x\right)\left(x^2-x+1\right)-6=0\)
Đặt \(x^2-x=t\Rightarrow t\left(t+1\right)-6=0\Rightarrow t^2+t-6=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x=-3\\x^2-x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x+3=0\left(vn\right)\\x^2-x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Bài 1:
1.
\((x^2-6x)^2-2(x-3)^2+2=0\)
\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x+9)+2=0\)
\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x)-16=0\)
Đặt $x^2-6x=a$ thì pt trở thành:
$a^2-2a-16=0$
$\Leftrightarrow a=1\pm \sqrt{17}$
Nếu $a=1+\sqrt{17}$
$\Leftrightarrow x^2-6x=1+\sqrt{17}$
$\Leftrightarrow (x-3)^2=10+\sqrt{17}$
$\Rightarrow x=3\pm \sqrt{10+\sqrt{17}}$
Nếu $a=1-\sqrt{17}$
$\Rightarrow x=3\pm \sqrt{10-\sqrt{17}}$
Vậy.........
2.
$x^4-2x^3+x=2$
$\Leftrightarrow x^3(x-2)+(x-2)=0$
$\Leftrightarrow (x-2)(x^3+1)=0$
$\Leftrightarrow (x-2)(x+1)(x^2-x+1)=0$
Thấy rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ nên $(x-2)(x+1)=0$
$\Rightarrow x=2$ hoặc $x=-1$
Vậy.......
Bài 2:
1.
ĐKXĐ: $x\neq 1$. Ta có:
\(x^2+(\frac{x}{x-1})^2=8\)
\(\Leftrightarrow x^2+(\frac{x}{x-1})^2+\frac{2x^2}{x-1}=8+\frac{2x^2}{x-1}\)
\(\Leftrightarrow (x+\frac{x}{x-1})^2=8+\frac{2x^2}{x-1}\)
\(\Leftrightarrow (\frac{x^2}{x-1})^2=8+\frac{2x^2}{x-1}\)
Đặt $\frac{x^2}{x-1}=a$ thì pt trở thành:
$a^2=8+2a$
$\Leftrightarrow (a-4)(a+2)=0$
Nếu $a=4\Leftrightarrow \frac{x^2}{x-1}=4$
$\Rightarrow x^2-4x+4=0\Leftrightarrow (x-2)^2=0\Rightarrow x=2$ (tm)
Nếu $a=-2\Leftrightarrow \frac{x^2}{x-1}=-2$
$x^2+2x-2=0\Rightarrow x=-1\pm \sqrt{3}$ (tm)
Vậy........
2. ĐKXĐ: $x\neq 0; 2$
$(\frac{x-1}{x})^2+(\frac{x-1}{x-2})^2=\frac{40}{49}$
$\Leftrightarrow (\frac{x-1}{x}+\frac{x-1}{x-2})^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$
$\Leftrightarrow 4\left[\frac{(x-1)^2}{x(x-2)}\right]^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$
Đặt $\frac{(x-1)^2}{x(x-2)}=a$ thì pt trở thành:
$4a^2-2a=\frac{40}{49}$
$\Rightarrow 2a^2-a-\frac{20}{49}=0$
$\Rightarrow a=\frac{7\pm \sqrt{209}}{28}$
$\Leftrightarrow 1+\frac{1}{x(x-2)}=\frac{7\pm \sqrt{209}}{28}$
$\Leftrightarrow \frac{1}{x(x-2)}=\frac{-21\pm \sqrt{209}}{28}$
$\Rightarrow x(x-2)=\frac{28}{-21\pm \sqrt{209}}$
$\Rightarrow (x-1)^2=\frac{7\pm \sqrt{209}}{-21\pm \sqrt{209}}$.
Dễ thấy $\frac{7+\sqrt{209}}{-21+\sqrt{209}}< 0$ nên vô lý
Do đó $(x-1)^2=\frac{7-\sqrt{209}}{-21-\sqrt{209}}$
$\Leftrightarrow x=1\pm \sqrt{\frac{7-\sqrt{209}}{-21-\sqrt{209}}}$
Vậy........
câu a:
\(8x^2-6x+3-2x=\left(2x-1\right)\sqrt{8x^2-6x+3}\)
đặt \(t=\sqrt{8x^2-6x+3}\Leftrightarrow t^2=8x^2-6x+3\)phương trình trở thành
\(t^2-2x=\left(2x-1\right)t\Leftrightarrow t^2-\left(2x-1\right)t-2x=0\)
có \(\Delta=\left(2x-1\right)^2+8x=\left(2x+1\right)^2\Rightarrow\orbr{\begin{cases}t=-1\\t=2x\end{cases}}\)
- \(t=-1\Rightarrow8x^2-6x+3=1\Leftrightarrow8x^2-6x+2=0VN\)
- \(t=2x\Rightarrow8x^2-6x+3=4x^2\Leftrightarrow4x^2-6x+3=0VN\)
Câu b:
Đặt \(t=\sqrt{x^2+1}\Leftrightarrow t^2=x^2+1\left(t>0\right)\)
PT\(\Leftrightarrow t^2-\left(x+3\right)t+3x=0\)
có :\(\Delta=\left(x+3\right)^2-4.3x=\left(x-3\right)^2\Rightarrow\orbr{\begin{cases}t=3\\t=x\end{cases}}\)
- \(t=3\Rightarrow9=x^2+1\Leftrightarrow x^2=8\Leftrightarrow\orbr{\begin{cases}x=2\sqrt{2}\\x=-2\sqrt{2}\end{cases}}\)
- \(t=x\Leftrightarrow x^2=x^2+1VN\)
a) Đặt x -3 = a
<=> a(a+2)(a+8)(a+10) - 297=0
<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0
<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)
Đặt \(a^2+10a=b\)
\(b^2+16b-297=0\)
\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)
b) bấm máy ra nhân tử chung :D
c)
\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)
\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
<=> x = 2018
d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)
giống câu c
\( a)2x - 3 = 3x - 7\\ \Leftrightarrow 2x - 3x = - 7 + 3\\ \Leftrightarrow - x = - 4\\ \Leftrightarrow x = 4\\ b)x - \left( {6 - 5x} \right) = 2\left( {x - 1} \right) + 12\\ \Leftrightarrow 6x - 6 = 2x - 2 + 12\\ \Leftrightarrow 6x - 2x = 10 + 6\\ \Leftrightarrow 4x = 16\\ \Leftrightarrow x = 4\\ c){x^4} - 144x = 2{x^2} + 1295\\ \Leftrightarrow {x^4} - 2{x^2} - 144x - 1295 = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {{x^3} - 5{x^2} + 23x - 259} \right) = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {x - 7} \right)\left( {{x^2} + 2x + 37} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 5\\ x = 7\\ {x^2} + 2x + 37 = 0\left( {vn} \right) \end{array} \right. \)
a) \(2x-3=3x-7\)
\(\Leftrightarrow x=4\)
b) \(x-\left(6-5x\right)=2\left(x-1\right)+12\)
\(\Leftrightarrow x-6+5x=2x-2+12\)
\(\Leftrightarrow\)\(4x=16\)
\(\Leftrightarrow x=4\)
c) \(x^4-144x=x^2+1295\)
\(\Leftrightarrow x^4-x^2-144x-1295=0\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(4x^2+144x+1295\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-\left(2x+36\right)^2=0\)
\(\Leftrightarrow\left(x^2+1+2x+36\right)\left(x^2+1-2x-36\right)=0\)
\(\Leftrightarrow\left(x^2+2x+37\right)\left(x^2-2x-35\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+36\right)\left(x^2+2x-7x-35\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+36\right]\left[\left(x+5\right)\left(x-7\right)\right]=0\)
do \(\left(x+1\right)^2+36\ge36\forall x\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=7\end{matrix}\right.\)
Vậy/...
=>x^4+2x^2+1-4x^2-144x-1296=0
=>(x^2+1)^2-(2x+36)^2=0
=>(x^2+1-2x-36)(x^2+1+2x+36)=0
=>x^2-2x-35=0
=>(x-7)(x+5)=0
=>x=7 hoặc x=-5