giải phương trình
\(\left(x-2,5\right)^4+\left(x-1,5\right)^4=1\)
giải các phương trình:
a)\(\left(x+3\right)^4+\left(x+5\right)^4=16\)
b)\(\left(x-2\right)^4+\left(x-3\right)^4=1\)
c)\(\left(x+1\right)^4+\left(x-3\right)^4=82\)
d)\(\left(x-2,5\right)^4+\left(x-1,5\right)^4=1\)
a) (x+3)4+(x+5)4=16
<=>(x+3)4+(x+5)4=04+24
TH1: \(\left\{{}\begin{matrix}x+3=0\\x+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=-3\end{matrix}\right.\Leftrightarrow x=-3\)
TH2:\(\left\{{}\begin{matrix}x+3=2\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)(loại)
b)(x-2)4+(x-3)4=1=04+14
TH1: \(\left\{{}\begin{matrix}x-2=0\\x-3=1\end{matrix}\right.\)loại
TH2: \(\left\{{}\begin{matrix}x-2=1\\x-3=0\end{matrix}\right.\)=>x=3.
c)(x+1)4+(x-3)4=82=34+(-1)4
làm tương tự => x=2.
d) làm tương tự câu b
Giải các phương trình sau:
a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\);
b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\);
c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\);
d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\).
a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\)
\(8 - x + 15 = 6 - 4x\)
\( - x + 4x = 6 - 8 - 15\)
\(3x = - 17\)
\(x = \left( { - 17} \right):3\)
\(x = \dfrac{{ - 17}}{3}\)
Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).
b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)
\( - 9 + 12u = - 45 + 6u\)
\(12u - 6u = - 45 + 9\)
\(u = \left( { - 36} \right):6\)
\(6u = - 36\)
\(u = - 6\)
Vậy nghiệm của phương trình là \(u = - 6\).
c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)
\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)
\({x^2} + 6x + 9 - {x^2} - 4x = 13\)
\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)
\(2x = 4\)
\(x = 4:2\)
\(x = 2\)
Vậy nghiệm của phương trình là \(x = 2\).
d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)
\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)
\({y^2} - 25 - {y^2} + 4y - 4 = 5\)
\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)
\(4y = 34\)
\(y = 34:4\)
\(y = \dfrac{{17}}{2}\)
Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).
Tìm \(x\in Q\):
\(c,\left|x-1,5\right|+\left|2,5-x\right|\) \(=0\)
\(d,\left|x-\dfrac{4}{5}\right|=\dfrac{3}{4}\)
\(\left[{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\left[{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=\dfrac{-3}{4}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)
c.x thuộc tập hợp rỗng
d. cậu chia ra thành 2 trường hợp nhé
Giải phương trình sau : \(\left|x-2\right|\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
TH1: \(x\ge2\)
\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=4\)
\(\Leftrightarrow x^4-5x^2=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\sqrt{5}\left(loại\right)\\x=\sqrt{5}\end{matrix}\right.\)
TH2: \(x< 2\)
\(-\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=-4\)
\(\Leftrightarrow x^4-5x^2+8=0\)
\(\Leftrightarrow\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm)
Vậy \(x=\sqrt{5}\)
Bài 1:
a) \(\left(x-1,3\right)^2=9\)
b) \(2^{4-x}=32\)
c) \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)
a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
b) 24-x = 32
⇔ 24-x = 25
⇔ 4-x=5
⇔ x=-1
c) (x+1,5)2+(y-2,5)10=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)
\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)
\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)
\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)
a. (x - 1,3)2 = 9
<=> (x - 1,3)2 - 9 = 0
<=> (x - 1,3)2 - 32 = 0
<=> (x - 1,3 - 3)(x - 1,3 + 3) = 0
<=> (x - 4,3)(x + 1,7) = 0
<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)
Giải phương trình:
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2\)
đkxđ: x khác 0
\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)
\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)
\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)
\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)
\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)
\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)
=> x=0 ( loại , ko tm)
Vậy pt vô nghiệm
giải phương trình :
\(\left(x-2\right)\left(x-1\right)\left(x-8\right)\left(x-4\right)=4x^2\)
\(\left(x-2\right)\left(x-1\right)\left(x-4\right)\left(x-8\right)=4x^2\)
\(\Leftrightarrow[\left(x-2\right)\left(x-4\right)][\left(x-1\right)\left(x-8\right)]=4x^2\)
\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-9x+8\right)=4x^2\)
thấy \(x=0;2\) không phải nghiệm của phương trình nên ta chia hai vế của pt cho \(x^2\) ta được \(:\)
\(\Leftrightarrow\left(x+\dfrac{8}{x}-9\right)\left(x+\dfrac{8}{x}-6\right)=4\)
\(Đặt:\) \(x+\dfrac{8}{x}=a\) thì pt trở thành \(:\)
\(\left(a-6\right)\left(a-9\right)=4\)
\(\Leftrightarrow a^2-15a+50=0\)
\(\Leftrightarrow\left(a-5\right)\left(a-10\right)=0\Leftrightarrow\left\{{}\begin{matrix}a=5\\a=10\end{matrix}\right.\)
\(Với\) \(a=5\) thì \(x+\dfrac{8}{x}=5\Leftrightarrow x^2-5x+8=0\left(vônghiem\right)\)
\(Với\) \(a=10\) thì \(x+\dfrac{8}{x}=10\Leftrightarrow x^2-10x+8=0\Leftrightarrow\left\{{}\begin{matrix}x=5-căn17\\x=5+căn17\end{matrix}\right.\)
\(Vậy...\)
Giải phương trình:
\(2\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
Phương pháp:
Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)
Thay vào pt
\(x\ne0:đặt:x+\dfrac{1}{x}=t\)
\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)
\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)
\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)
\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)
(bn xem lại đề)
Giải phương trình \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\dfrac{-3x^2}{4}\)
PT tương đương
\(\left(x^2+7x+6\right)\left(x^2+5x+6\right)=\dfrac{-3x^2}{4}\)
Xét \(x=0\Rightarrow6.6=0\)(vô lý)
Xét \(x\ne0\). Ta chia 2 vế của PT cho \(x^2\ne0\). PT tương đương
\(\left(x+\dfrac{6}{x}+7\right)\left(x+\dfrac{6}{x}+5\right)=\dfrac{-3}{4}\)
Đặt \(x+\dfrac{6}{x}+5=t\)
PT\(\Leftrightarrow t\left(t+2\right)=\dfrac{-3}{4}\Leftrightarrow t^2+2t+1=\dfrac{1}{4}\)
\(\Leftrightarrow\left(t+1\right)^2=\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}t+1=\dfrac{-1}{2}\\t+1=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-3}{2}\\t=\dfrac{-1}{2}\end{matrix}\right.\)
Đến đây bạn thay vào là tìm được nghiệm nhé.