Tim x biet
\(x^4-2x^3-2x^2+3x+2=0\)
Những câu hỏi liên quan
Tim xthuoc Z biet:
1,|2x-5|-|2x+9|=0
2,|x+1|-|x+2|-|3-x|=7
3,|2x+3|+|3x+2|-|4-x|=10
bai1.tim x biet:
a,(x+2).(x+3)-(x-2).(x+5)=0
b,(2x+3).(x-4)+(x-5).(x-2)=(3x-5).(x-4)
c,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)=33
,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn
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Xem thêm câu trả lời
tim x biet:
x+x2-x3-x4=0
2x3+3x2+2x2+3=0
x2-x-12=0
a)x+x2-x3-x4=0
<=>x(x+1)-x3(x+1)=0
<=>x(x+1)(1-x2)=0
<=>x(x+1)(x+1)(x-1)=0
<=>x(x+1)2(x-1)=0
<=>x=0
hoặc (x+1)2=0<=>x=-1
hoặc x-1=0<=>x=1
b)sửa đề 1 chút!!!
2x3+3x2+2x+3=0
<=>x2(2x+3)+(2x+3)=0
<=>(2x+3)(x2+1)=0
<=>2x+3=0(do x2+1>0 với mọi x)
<=>2x=-3
<=>x=-1,5
c)x2-x-12=0
<=>(x2-4x)+(3x-12)=0
<=>(x(x-4)+3(x-4)=0
<=>(x-4)(x+3)=0
<=>x-4=0<=>x=4
Hoặc x+3=0<=>x=-3
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ngu có thế cũng ko biết
tim x biet x2-11x+18=0
-4x2+5x-1=0 2x3+3x2+2x+3=0x(2x-7)-4x+14=0tinh nhanh bt sau
A=3.(x-3).(x+7)+(x-4)2+48,tai x=0.5
a) x^2 - 11x + 18 = 0
=> x^2 - 2x - 9x + 18 = 0
=> x ( x- 2 ) - 9 ( x- 2 ) = 0
=> ( x- 9 )( x- 2 )= 0
=> x- 9 = 0 hoặc x - 2 = 0
=> x= 9 hoặc x = 2
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Tim x biet rang:
a) 4/4^x + 3.4^2-x = 832
b) (3x - 2)^2002 = (3x - 2)^2004
c) /1 - 2x/ + x + 2 = 0
tim x bieta/(3x-5)(2x-1)-(x+2)(6x-1)0b/ (3x-5)(3x+2)-(3x-1)2-5c/(3x+2)(x-5)3(x-1)2-2d/ (x+1)2/3 - (x-2)2/2 2x+1/2 (x-3)2/6g/49x2(3x+2)2h/(3x-4)2-(2x-2)2-3(x-2)(2x-1)0i/ (x-2)(x2-2x+4)-x(x2+2)15k/ 6x2-7x-30m/(x+5)(x-3)+x2-250e/ x3+3x24x+12f/ (6x+7)2(3x+4)(x+1)6
Đọc tiếp
tim x biet
a/(3x-5)(2x-1)-(x+2)(6x-1)=0
b/ (3x-5)(3x+2)-(3x-1)2=-5
c/(3x+2)(x-5)=3(x-1)2-2
d/ (x+1)2/3 - (x-2)2/2 = 2x+1/2 (x-3)2/6
g/49x2=(3x+2)2
h/(3x-4)2-(2x-2)2-3(x-2)(2x-1)=0
i/ (x-2)(x2-2x+4)-x(x2+2)=15
k/ 6x2-7x-3=0
m/(x+5)(x-3)+x2-25=0
e/ x3+3x2=4x+12
f/ (6x+7)2(3x+4)(x+1)=6
2) tim x biet
a) (3x-5)2-(x+1)2=0
b) (5x-4)2-49x2=0
c) 4x3-36x=0
d) (2x+3) (r-1)+(2x-3) (1-x)=0 giai gium minh
a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy ...
b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy ...
d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy ...
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Tim x biet:
A. (x-3/4).(3x+1/2)lon hon hoac bang 0
B. (2x+1).(4x+3)be hon hoac bang 0
tim x biet (2x-2)*(3x+6)=0
\(\left(2x-2\right).\left(3x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2=0\\3x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\3x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)
ĐS: ...........
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( 2x - 2 ) . ( 3x + 6 ) = 0
=> 2x - 2 = 0 hoặc 3x + 6 = 0
2x = 0 + 2 hoặc 3x = 0 - 6
2x = 2 hoặc 3x = - 6
x = 1 hoặc x = - 2
*Lưu ý nếu bạn học số nguyên rồi thì mới làm theo cách này nha!
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