tính
a)\(101^2-99^2\)
b)98.102
c)\(77^2+23^2+77.46\)
d)M=\(x^3+9x^2+27x+27\)tại x=7
TÌM X
a)\(2x^2+3x-5=0\)
b)\(2x^2-11x-51=0\)
AI GIÚP MK VS MAI MK KIỂM TRAI RÙI
B=x^3+3x^2+3x+1 với x=99
C=x^3-9x^2+27x-26 với x=23
D=(2x-3)^2-(4x-6)(2x-5)+(2x-5) với X= 99
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Bài 5 : Tìm x, biết :
a, x^2 ( x-5 ) + 5 - x = 0
b, 3x^4 - 9x^3 = -9x^2 + 27x
c, x^2 ( x+8 ) + x^2 = -8x
d, ( x+3 ) ( x^2 - 3x + 5 ) = x^2 + 3x
Giúp mk vs ạ mk đang cần gấp
a/ \(x^2\left(x-5\right)+5-x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
Vậy...
b/ \(3x^4-9x^3=-9x^2+27x\)
\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)
Vì \(x^2+3>0\forall x\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy..
c/ \(x^2\left(x+8\right)+x^2=-8x\)
\(\Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\)
\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)
\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)
Vậy...
d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)^2+1\right]=0\)
Vì \(\left(x-2\right)^2+1>0\forall x\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy..
\(a,x^2\left(x-5\right)+5-x=0\\ \Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
\(b,3x^4-9x^3=-9x^2+27x\\ \Leftrightarrow3x^4-9x^3+9x^2-27x=0\\ \Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\\ \Leftrightarrow3x\left(x-3\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
do \(x^2+2>0\)
\(c,x^2\left(x+8\right)+x^2=-8x\\ \Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\\ \Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\\ \Leftrightarrow x\left(x+1\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-8\end{matrix}\right.\)
\(d,\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x^2-3x=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5-x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-4x+5=0\end{matrix}\right.\)
\(\Delta=b^2-4ac=\left(-4\right)^2-4\cdot1\cdot5=16-20=-4< 0\)\(\rightarrow\)\(x^2-4x+5\) vô nghiệm
Vậy \(x=-3\)
mai mik kiểm tra rùi giúp mik vs pls
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
1.Tìm x:
a,x mũ 3 - 16 = 0
b,x mũ 4 - 2x mũ 3 + 10x mũ 2 - 20x = 0
c,(2x - 3)mũ 2 = (x + 5)mũ 2
d,x mũ 2(x - 1) - 4x mũ 2 + 8x -4 = 0
e,x mũ 3 - 11x mũ 2 + 30x = 0
P/s:Giúp mk vs chiều mai mk phải nộp rồi
b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
=>x=1 hoặc x=2
Tìm x bt
1) 3.(2x−1).(3x−1)−(2x−3).(9x−1)−3=−3
2) 3x−1.(2x+7)−x+1.(6x−5)=x+2−(x−5)
3) 3xy.(x+y)−(x+y).(x2+y2+2xy)+y3=27
4) 5.x+1.(1−x).(2x2+3)=0
giúp mk vs mai mk đi hk rùi
Bài 6 : Tìm x, biết :
a, ( 2x-5)^2 - ( 5+2x)^2 = 0
b, 27x^3 - 54x^2 + 36x = 8
c, ( x^3 + 8 ) - ( x+2) ( x-4) = 0
d, x^6 - 1 = 0
Giúp mk vs ạ mk đang cần gấp
Câu 1: Tính giá trị các biểu thức:
a. A= 4x^2 + 8x + 5 với x = 49
b. B= x^3 + 3x^2 + 3x + 1 với x = 99
c. C= x^3 - 9x^2 + 27x - 26 với x = 23
d. D= (2x - 3)^2 - (4x - 6)(2x - 5) + (2x - 5)^2 với x = 99
Mng ơi giúp mk vs ak mk can gap ak:
a) x^3-10x-3=0
b) -2x^3-9x^2-2x+1=0
c) 2x^4+3x^3-5x^2-7x-2=0
d) x^4-2x^3-3x^2+4x+3=0(*)