Bài 1: Tìm x
a) (x-1)^3+3(x+1)^2=(x^2-2x+4)(x+2)
b) x^2-4=8(x-2)
c) x^2-4x+4=9(x-2)
d) 4x^2-12x+9=(5-x)^2
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: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
a) (x - 1)3 + 3(x + 1)2 = (x2 - 2x + 4)(x + 2)
b) x2 - 4 = 8(x - 2)
c) x2 - 4x + 4 = 9(x - 2)
d) 4x2 - 12x + 9 = (5 - x)2
a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+2x+1=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+2x-x^3=1-1+8\)
\(\Leftrightarrow5x=8\)
\(\Leftrightarrow x=\dfrac{8}{5}\)
Vậy \(S=\left\{\dfrac{8}{5}\right\}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-6=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-6=0\Leftrightarrow x=6\)
Vậy \(S=\left\{2;6\right\}\)
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow x-2=0\) hoặc \(x-11=0\)
:) \(x-2=0\Leftrightarrow x=2\)
:) \(x-11=0\Leftrightarrow x=11\)
Vậy \(S=\left\{2;11\right\}\)
(d ko bít lèm)
#IDOL
d, \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Rightarrow4x^2-12x+9=25-10x+x^2\)
\(\Rightarrow4x^2-x^2-12x+10x+9-25=0\)
\(\Rightarrow3x^2-2x-16=0\)
\(\Rightarrow\left(-2-x\right)\left(8-3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2-x=0\\8-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=2\\-3x=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Tìm x
a, 3(x-1)^2-3x(x-5)=2
b, 4x^2-12x=-9
c, (2x-3)^2=(x+5)^2
d, (x^4-2x^3+4x^2-8x)÷(x^2+4)-2x=-4
e, x-2/2-x+3/3+x+4/5-x+5=0
\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)
\(3x^2-6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
\(x=-\frac{1}{9}\)
\(b.4x^2-12x+9=0\)
\(4x^2-6x-6x+9=0\)
\(2x\left(x-3\right)-3\left(x-3\right)=0\)
\(\left(2x-3\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
a) 3(x - 1)2 - 3x(x - 5) = 2
=> 3(x2 - 2x + 1) - 3x2 + 15x = 2
=> 3x2 - 6x + 3 - 3x2 + 15x = 2
=> 9x = 2 - 3
=> 9x = -1
=> x = -1/9
b) 4x2 - 12x = -9
=> 4x2 - 12x + 9 = 0
=> (2x - 3)2 = 0
=> 2x - 3 = 0
=> 2x = 3
=> x = 3/2
c) (2x - 3)2 = (x + 5)2
=> (2x - 3)2 - (x + 5)2 = 0
=> (2x - 3 - x - 5)(2x - 3 + x + 5) = 0
=> (x - 8)(3x + 2) = 0
=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
d) \(\left(x^4-2x^3+4x^2-8x\right):\left(x^2+4\right)-2x=-4\)
=> \(\left[x^3\left(x-2\right)+4x\left(x-2\right)\right]:\left(x^2+4\right)-2x=-4\)
=> \(x\left(x-2\right)\left(x^2+4\right):\left(x^2+4\right)-2x=-4\)
=> \(x^2-2x-2x+4=0\)
=> \(\left(x-2\right)^2=0\)
=> x - 2 = 0
=> x = 2
e) khđ
ai giúp mik với mik nhiều bt lắm
a, 1/x-1 - 7/x+2 = 3/x2+x-2
b,x+3/x-4 + x-1/x-2 = 2/6x-8-x2
c,1/x+1 + 2/x3-x2x+1 = 3/1-x2
d, 3x-1/x-1 - 2x+5/x+3 = 1 - 4/x2+2x-3
e, x2+2x+1/x2+2x+2 + x2+2x+2/x2+2x+3 = 7/6
f, 1/4x2-12x+9 - 3/9-4x2=4/4x2+12x+9
Tìm x:
a, (x-1)3+3(x+1)2=(x2-2x+4)(x+2)
b, x2-4=8(x-2)
c, x2-4x+4=9(x-2)
d, 4x2-12x+9=(5-x)2
a,\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x-1+3x^2+6x+3=8\)
\(\Leftrightarrow9x=6\)
\(\Leftrightarrow x=\frac{2}{3}\)
b,\(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2+12x-8x=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
c,\(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow x^2-4x+4=9x-18\)
\(\Leftrightarrow x^2-4x+4-9x+18=0\)
\(\Leftrightarrow x^2-13x+22=0\)
\(\Leftrightarrow x^2-2x-11x+22=0\)
\(\Leftrightarrow x\left(x-2\right)-11\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
d,\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10-x^2=0\)
\(\Leftrightarrow3x^2-2x-16=0\)
\(\Leftrightarrow3x^2+6x-8x-16=0\)
\(\Leftrightarrow3x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\)
bài 1 :1)2/x-1 + 2x+3/x^2+x+1=(2x+1)(2x-1)/x^3-1
2)x^3-(x+1)^3/(4x+3)(x-5)=7x-1/4x+3 - x/x-5 (x=-1/9)
3)12/1-9x^2=1-3x/1+3x - 1+3x/1-3x (x=-1)
4)x+5/x-1=x+1/x-3 - 8/x^2-4x+3
5)1/x-1 + 2x/x+3=-1 (x=0,-1/3)
6)1/3y^2-10y+3=6y/9y^2-1 + 2/1-3y (y=1)
7)24/x^2-2x+4=3x/x+2 + 72/x^3+8 (x=2)
8)1/x^2+9x+20 +1/x^2+11x+30 +1/x^2+13x+42=1/18 (-13,2)
9)x+4/2x^2-5x+2 + x+1/2x^2-7x+3=2x+5/2x^2-7x+3 (x=4)
10)12x/x-4 - 3x^2/x+4=384/x^2-16
bài 2:
tìm giá trị lớn nhất và nhỏ nhất của các đa thức sau
A=x^2+4x+5 B=-x^2-2x+2 C= x^2+2x+3 D=-x^2+4x+2000
E=10x-4x^2-23 F=1/x^2-2x+3 G=3x^2+3x+5/x^2+x+1 H=x^2+x+1/x^2-x+1
O=5x^2+8xy+5y^2 P=42-x/x-15
bài 3: so sánh A và B biết : A=2003.2005 và 2004^2
Tìm GTLN - GTNN của các biểu thức ?
* bài 1: Tìm GTNN:
a) A= (x - 5)² + (x² - 10x)² - 24
b) B= (x - 7)² + (x + 5)² - 3
c) C= 5x² - 6x +1
d) D= 16x^4 + 8x² - 9
e) A= (x + 1)(x - 2)(x - 3)(x - 6)
f) B= (x - 2)(x - 4)(x² - 6x + 6)
g) C= x^4 - 8x³ + 24x² - 8x + 25
h) D= x^4 + 2x³ + 2x² + 2x - 2
i) A= x² + 4xy + 4y² - 6x – 12y +4
k) B= 10x² + 6xy + 9y² - 12x +15
l) C= 5x² - 4xy + 2y² - 8x – 16y +83
m) A= (x - 5)^4 + (x - 7)^4 – 10(x - 5)²(x - 7)² + 9
* Bài 2: Tìm GTLN:
a) M= -7x² + 4x -12
b) N= -16x² - 3x +14
c) M= -x^4 + 4x³ - 7x² + 12x -5
d) N= -(x² + x – 2) (x² +9x+18) +27
* Bài 3:
1) Cho x - 3y = 1. Tìm GTNN của M= x² + 4y²
2) Cho 4x - y = 5. Tìm GTNN của 3x²+2y²
3) Cho a + 2b = 2. Tìm GTNN của a³ + 8b³
* Bài 4: Tìm GTLN và GTNN của các biểu thức:
1) A = (3 - 4x)/(x² + 1)
2) B= (8x + 3)/(4x² + 1)
3) C= (2x+1)/(x²+2)
Toán lớp 1 cái gì,xạo.Toán trung học thì có.
Lớp 1 mà làm được cái này thì...THIÊN TÀI
1) tìm x :
a) (x-3)(x+3)-(x+5)(x-1)=6
b)(3x-2)^2-(2x+1)^2-(5x-1)(x+1)=20
c)(2x+1)(4x^2-2x+1)+(3-2x)(9+6x+4x^2)+12x=4
a) x^4+2x^3-2x -1
b)9+ 6x+ x^2+x^4
c) a^3+b^3+c^3-3abc
d)x^4-4x^3-2x^2+12x+9
e)x^3+3x^2-x-3
f) x^3+2x^2+x+1
g) x^3+3x^2+x+2
h)x^3+5x^2+x+9
i) x^6-9x^3+8
k) x^5+x^4+1
l) x^7+x^5+ 1