giải phương trình sau:
a, \(x^4+x^2-2=0\)
b,\(x^4-13x^2+36=0\)
c, \(\dfrac{1}{8}x^3+\dfrac{3}{4}x^4+\dfrac{3}{2}x+=-1\)
giải các phương trình sau
a, 3(x-1) -3=2(x+3)
b, \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)
c,\(\left(2x-1\right)^2-x^2=0\)
d,\(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)
\(\Leftrightarrow3x-3-3=2x+6\)
\(\Leftrightarrow x=12\)
b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)
\(\Leftrightarrow3x+12-4x-12=2x+12\)
\(\Leftrightarrow-3x=12\)
hay x=-4
c: Ta có: \(\left(2x-1\right)^2-x^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
Giải các bất phương trình sau
a) (x-4)2<x(x-8)
b) x+\(\dfrac{1}{2}\)\(\overset{>}{-}\)\(\dfrac{3-5x}{-3}\)
c) \(\dfrac{x-7}{-4}\)\(\overset{< }{-}\)\(\dfrac{4-2x}{-3}\)
a: =>x^2-8x+16<x^2-8x
=>16<0(loại)
b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)
=>x+1/2>=5/3x-1
=>-2/3x>=-3/2
=>x<=3/2:2/3=9/4
c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)
=>21-3x<=8x-16
=>-11x<=-37
=>x>=37/11
giải các phương trình sau
a, 3x -(3x+2) =x+3
b, \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
c, \(\left(x^2-3^2\right)+2\left(x-3\right)=0\)
d,\(\dfrac{1}{x-1}+\dfrac{2}{1+x}-\dfrac{4x+6}{x^2-1}=0\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
giải phương trình :
a, \(2x^2-11x+21-3\sqrt[3]{4x-4}=0\)
b, \(\left(3x-2\right)\sqrt{x+1}-x^2-x-2=0\)
c, \(x+4-2\left(\dfrac{x+2}{x-1}\right)\sqrt{\dfrac{x-1}{x+2}}=0\)
c.
ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)
\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)
\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))
\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)
\(\Rightarrow x^3+7x^2+4x-24=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)
a.
\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)
Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)
Ta có:
\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
b.
ĐKXD: \(x\ge-1\)
Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)
Đặt \(\sqrt[]{x+1}=t\ge0\)
\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)
\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Bài 1: Giải phương trình
\(a,\dfrac{x+1}{2009}+\dfrac{x+3}{2007}=\dfrac{x+5}{2005}+\dfrac{x+7}{1993}\)
\(b,\left(x+2\right)^4+\left(x+4\right)^4=14\)
\(c,\left(x-3\right)\left(x-2\right)x+1=60\)
d, \(2x^4+3x^3-x^2+3x+2=0\)
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
c)Ta có: \(\left(x-3\right)\left(x-2\right)\left(x+1\right)=60\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x+1\right)=60\)
\(\Leftrightarrow x^3+x^2-5x^2-5x+6x+6-60=0\)
\(\Leftrightarrow x^3-4x^2+x-54=0\)
Bạn xem lại đề, nghiệm rất xấu
giải phương trình sau
\(a,\dfrac{4x-17}{2x^2+1}=0\\ b,\dfrac{4}{x-2}-x+2=0\\ c,x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\\ d,\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=3^1_5\)
a) \(\dfrac{4x-17}{2x^2+1}=0\) ( 1)
ĐKXĐ : \(2x^2+1\ne0\)
⇔ x2 \(\ne\dfrac{-1}{2}\) ( luôn đúng )
( 1) ⇔ 4x - 17 = 0
⇔ x = \(\dfrac{17}{4}\)
KL....
b) \(\dfrac{4}{x-2}-x+2=0\)
⇔ \(\dfrac{4-\left(x-2\right)^2}{x-2}=0\left(x\ne2\right)\)
⇔ 4 - ( x2 - 4x + 4) = 0
⇔ 4x - x2 = 0
⇔ x( 4 - x) = 0
⇔ x = 0 ( TM) hoặc x = 4 ( TM)
KL...
c) x + \(\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\left(x\ne0\right)\)
Đặt : x + \(\dfrac{1}{x}=t\) ⇒ \(x^2+\dfrac{1}{x^2}=t-2\) , ta có :
t = t - 2 ( Vô lý )
⇒ t ∈ ∅
⇒ Phương trình vô nghiệm
d) \(\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=3\dfrac{1}{5}\left(x\ne2;x\ne4\right)\)
⇔ ( x - 3)( x - 4) - ( x - 2)2 = \(\dfrac{16}{5}\)( x - 2)( x - 4)
⇔ x2 - 7x + 12 - x2 + 4x - 4 = \(\dfrac{16}{5}\)( x2 - 6x + 8)
⇔ \(\dfrac{40-15x}{5}=\dfrac{16x^2-96x+128}{5}\)
giải nốt nha
Mạn phép ko chép lại đề mk làm lại câu c
Đặt : t = x + \(\dfrac{1}{x}\) ⇒ t2 - 2 = x2 + \(\dfrac{1}{x^2}\) , ta có :
t = t2 - 2
⇔ t2 - t - 2 = 0
⇔ t2 + t - 2t - 2 = 0
⇔ t( t + 1) - 2( t + 1) = 0
⇔ ( t + 1)( t - 2) = 0
⇔ t = 2 hoặc t = -1
* Với t = 2 ,ta có :
x + \(\dfrac{1}{x}=2\)
Tự giải ra nha
* Với t = -1 , ta có :
x + \(\dfrac{1}{x}=-1\)
Tự giải ra nha
Giai phương trình :
a)\(\dfrac{2x-1}{3}-x=\dfrac{x+3}{4}+2\)
b)\(x^2-4+\left(x-9\right)\left(x-2\right)=0\)
c)\(\dfrac{x-1}{x-3}-\dfrac{1}{x+3}=\dfrac{3x+3}{x^2-9}\)
a: =>4(2x-1)-12x=3(x+3)+24
=>8x-4-12x=3x+9+24
=>-4x-4=3x+33
=>-7x=37
=>x=-37/7
b: =>(x-2)(x+2+x-9)=0
=>(2x-7)(x-2)=0
=>x=2 hoặc x=7/2
c: =>(x-1)(x+3)-x+3=3x+3
=>x^2+2x-3-x+3=3x+3
=>x^2+x-3x-3=0
=>x^2-2x-3=0
=>(x-3)(x+1)=0
=>x=-1