tìm x biết
e, ( 9x\(^2\)- 49)+ (3x + 7 ) ( 7x+3) = 0
g, ( x - 4)\(^2\)= ( 2x-1)\(^2\)
h, x\(^2\)- x\(^2\)+ x - 1 = 0
i, x + x\(^2\)+ x + 1 = 0
k, x(x+ 16) - 7x - 42 = 0
m, x\(^2\)+ 7x + 12 = 0
n, x\(^2\)- 7x + 12 = 0
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
\(m,x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
\(n,2x^2+5x-3=0\)
\(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(k,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-4x-7x+14=0\)
\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
bài 7
4x3 + 12 = 120
b, ( x - 4 )2 = 64
c, ( x + 1 )3 - 2 = 52
d, 136 - ( x + 5)2 = 100
e, 4x = 16
f, 7x. 3 - 147 = 0
g, 2x+3 - 15 = 17
h, 52x-4. 4 = 102
i, (32 - 4x)(7 - x) = 0
k, ( 8 - x)(10 - 2x) = 0
m, 3x + 3x+1 = 108
n, 5x+2 + 5x+1 = 750
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
1) (x^2+x)^2-2(x^2+x)-15
2) (x^2+2x)^2+9x^2+18x+20
3) (x^2+3x+1)(x^2+3x+2)-6
4) x^2+2xy+y^2-x-y-12
5) x^2-16-4xy+4y^2
6) x^4 -3x^3-x+3
7) x^2-10x+16
8) 2x^2+7x+3
9) 3x^2+7x+2
10) 2x^2+5x-3
tìm x
a, \(x^2-4x+\dfrac{1}{4}\)
b, \(8x^2-25\)
c, \(x^2+7x=8\)
d, \(x^3-3x=-27+9x\)
e, x(x-3)-7x=-21
f, \(x^3-2x^2+x-2=0\)
g, \(x^2-4x=-4\)
h, \(x^3-x^2+x=-1\)
\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
a, 3y-2y=2y-3
3y-2y-2y=3
-y=3
y=-3
b, 3-4x+24+6x=x+27+3x
-4x+6x-x-3x =27-3-24
-2x =0
x =0
c, 5-(6-x)=4.(3-2x)
5-6+x =12-8x
x+8x =12+6-5
9x =13
x =13/9
d, 4.(x+3)=-7x+17
4x+12 =-7x+17
4x+7x =17-12
11x =5
x =5/11
A) \(3y-2y=2y-3\) \(\Leftrightarrow y-2y=-3\) \(\Leftrightarrow y=3\)
B) \(3-4x+24+6x=x+27-3x\)\(\Leftrightarrow3-4x+24+6x-x-27-3x=0\)
\(\Leftrightarrow-2x=0\)\(\Leftrightarrow x=0\)
C) \(5-\left(6-x\right)=4\left(3-2x\right)\)\(\Leftrightarrow5-6x+x-12+8x=0\)\(\Leftrightarrow9x-13=0\)\(\Leftrightarrow x=\frac{13}{9}\)
D) \(4\left(x+3\right)=-7x+17\) \(\Leftrightarrow4x+12+7x-17=0\)\(\Leftrightarrow11x-5=0\)\(\Leftrightarrow x=\frac{5}{11}\)
E) \(11x+42-2x=100-9x-22\)\(\Leftrightarrow11x+42-2x-100+9x+22=0\)
\(\Leftrightarrow18x-36=0\)\(\Leftrightarrow x=2\)
F) \(5x+\frac{3}{12}=1+\frac{2x}{9}\)\(\Leftrightarrow180x+9-36-8x=0\)\(\Leftrightarrow172x-27=0\)\(\Leftrightarrow x=\frac{27}{172}\)
TK nka !!! Mk giải tiếp !!
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
c)3x^2-7x-10=0
d)2x(x-10)-x+10=0
e)3x^3+7x^2+17x+5=0
f)(2x-1)^2-(x-3)^2=0
g)x^3-5x^2+8x=4
c, \(3x^2-7x+10=0\)
\(\Leftrightarrow3x^2+3x-10x+10=0\)
\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)
d, \(2x\left(x-10\right)-x+10=0\)
\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)
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
giải các phương trình sau:
a.2x + 3 = 7x - 7
b. \(\dfrac{x}{x+2}+\dfrac{x-1}{x-2}=\dfrac{2x^2+x}{x^2-4}\)
c.|x-1|=1
d.6x - 2 = 2x + 10
e. \(\dfrac{x}{2}+\dfrac{x+1}{3}=\dfrac{5}{2}\)
f.|x-1|=x+2
g.9x - 6 = 3x+12
h.x(x-1)+3(x-1)=0
i.\(\dfrac{7}{x}+\dfrac{2}{x+1}=\dfrac{x+23}{x\left(x+1\right)}\)
j.8x - 3=6x +9
k. |x-1|-4=5
l.\(\dfrac{x-5}{x^2-16}+\dfrac{3}{x+4}=\dfrac{7}{x-4}\)
a)2x + 3 = 7x - 7
(=)2x-7x=-7-3
(=)-5x=-10
(=)x=-2
Vậy S={2}