\(x\left(x-4\right)-\)(x mũ 2-8)=0
Tìm x,biết
a)\(\left(x-2^2\right)-1=0\)
b)\(4-\left(x-2\right)^2=0\)
c)\(x^2-9-\dfrac{8}{9}x^2=0\)
d)\(\left(3x-2\right)^2-\left(2x+3\right)^2=5\left(x+4\right)\left(x-4\right)\)
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
a)\(\left(x-2^2\right)-1=0\Rightarrow x-4-1=0\Rightarrow x=5\)
Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Giải các phương trình sau
1. \(\left(x-1\right)\left(x+5\right)\left(x^2+4x+8\right)+40=0\)
2. \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15=0\)
Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
Giải các phương trình sau:
a \(x^4-x^2-56=0\)
b \(\left(x-2\right)^4+\left(x+2\right)^4=32\)
c \(\left(x+3\right)^4+\left(x+5\right)^4=16\)
d \(\left(6-x\right)^4+\left(8-x\right)^4=80\)
a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Vậy x = 8 hoặc x = -7
a: Ta có: \(x^4-x^2-56=0\)
\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)
\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow x^2-8=0\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
Biết x0 là một nghiệm của phương trình: \(x^4-\left(2+\sqrt{8}\right)x^2-\left(3+\sqrt{8}\right)=0\). Tính giá trị của biểu thức A=\(x^6-\left(2+2\sqrt{8}\right)x^4+\left(678+\sqrt{8}\right)x^2+8-670\sqrt{8}\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2+y^2\right)+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+y^2+\left(x+y-2\right)\left(y+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2+y^2-4\right)\left(y+2\right)=-x\left(x^2+y^2\right)\\-\left(x^2+y^2\right)=\left(x+y-2\right)\left(y+2\right)\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2-4\right)\left(y+2\right)=x\left(x+y-2\right)\left(y+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\)
\(\Rightarrow x^2+y^2-4=x^2+x\left(y-2\right)\)
\(\Leftrightarrow\left(y+2\right)\left(y-2\right)=x\left(y-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\x=y+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết
Giải các phương trình sau:
a) \(x^2+\dfrac{2x}{x-1}=8\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
c) \(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)
d) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)=24\)
e) \(\left(x^2+x-2\right)\left(x^2+9x+18\right)=28\)
f) \(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4=0\)
g) \(x^4+6x^3+11x^2+6x+1=0\)
h) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
i) \(\left(x+2\right)^4+\left(x+8\right)^4=272\)
giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn
a) x2+\(\dfrac{2x}{x-1}\)=8(ĐKXĐ : x ≠ 1
⇔ x2(x-1)+2x =8⇔ x3 - x2 +2x - 8=0
⇔x3 - 23 -x2+2x =0⇔ (x-2)(x2 +x+1) -x(x-2)
⇔(x-2)(x2 +1)⇒x =2
x2 +1 =0⇒x2 -1⇒x ∈∅(loại)
vậy x =2
Giải phương trình :
a) \(x^2+\dfrac{2x}{x-1}=8\)
ĐKXĐ : \(x-1\ne0\Rightarrow x\ne1\)
Ta có : \(x^2+\dfrac{2x}{x-1}=8\)
\(\Leftrightarrow\) \(\dfrac{x^2\left(x-1\right)}{x-1}+\dfrac{2x}{x-1}=\dfrac{8\left(x-1\right)}{x-1}\)
\(\Rightarrow x^2\left(x-1\right)+2x=8\left(x-1\right)\)
\(\Leftrightarrow x^3-x^2+2x=8x-8\)
\(\Leftrightarrow x^3-x^2+2x-8x=-8\)
\(\Leftrightarrow x^3-x^2-6x+8=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x-8\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-2\left(3x-4\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)
Đối chiếu với ĐKXĐ ta được \(x\in\left\{\sqrt{2};\dfrac{4}{3}\right\}\) thỏa mãn.
rút gọn
C=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right)\div\dfrac{\sqrt{x}}{x-4}vớix>0,x\ne4\)
D=\(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x+1}}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}vớix>1,x\ne4,x\ne9\)
lm nhanhgiups mk nhé!Mk đang cần gấp!
c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)