b)-5 +/3x-1/+6=/-4/
c)(x-1)^2 = (x-1)^4
d) 5^-1 .25^x =125 (x thuộc Z)
e) /x+1/ +/x+2/+/x+3/=4x
Tìm x biết:a) 10/x = -15/9; b) x/9 = -11/5 : 0,6; c) -7/8 - 2x = -3/4; d) (x-1/2):1/3+5/7=9và5/7; e)1/15.x+4/5.x=5và1/5; f) -1/3<x/6<1/2(x thuộc Z); h) 3/5+2/5:x=-1/4;i) 4 và 3/4x - 3 và 1/2=5/4; k)9/4.(1/3x-1/2)=4và1/2
1)
a) 2-x/2001 - 1=1-x/2002 - x/2003
b)x^3 + 3x^2 + x + 3=0
c)/x-4/=/3-2x/
d)4X^2 + 16x +17
e)13-2/3x+2/=-1
f)/3x-4/=x-5
2)
a) tìm x thuộc Z để A=3x^2 - 9x + 2/x-3 thuộc Z
b)với giá trị nào của n thuộc Z thì A=3n+9/n-4 thuộc Z
3) chứng minh các bất phương trình sau vô nghiệm
a)x^2+x+2 nhỏ hơn bằng 0 b)x^4-2x^2+5 nhỏ hơn bằng 0
4)
1) x^4-8x^3+11x^2+8x-12=0 2)-3x^4+20x^3-35x^2-10x+48=0
3)x^5-5x^4+6x^3-x^2+5x-6=0 4)(x^2+x+1)(x^2+x+2)=12
5)(x-3)(x-5)(x-6)(x-10)-24x^2=0 6)(x+1)(x-4)(x+2)(x-8)+4x^2=0
7)(x^2-4)(x^2-10)=72
a) \(-5+|3x-1|+6=|-4|\)
b)\(\left(x-1\right)^2=\left(x-1\right)^4\)
c)\(5^{-1}.25^x=125\left(x\in Z\right)\)
d)\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
a) -5 + |3x - 1| + 6 = |-4|
=> -5 + |3x - 1| + 6 = 4
=> 1 + |3x - 1| = 4
=> |3x - 1| = 4 - 1
=> |3x - 1| = 3
=> \(\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)
=> \(\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy ...
d) |x + 1| + |x + 2| + |x + 3| = 4x
Ta có: |x + 1| \(\ge\)0 \(\forall\)x
|x + 2| \(\ge\)0 \(\forall\)x
|x + 3| \(\ge\)0 \(\forall\)x
=> |x + 1| + |x + 2| + |x + 3| \(\ge\)0 \(\forall\)x => 4x \(\ge\)0 \(\forall\) x=> x \(\ge\)0 \(\forall\)x
=> x + 1 + x + 2 + x + 3 = 4x
=> 3x + 6 = 4x
=> 6 = 4x - 3x
=> x = 6
Vậy...
b) (x - 1)2 = (x - 1)4
=> (x - 1)2 - (x - 1)4 = 0
=> (x - 1)2 .[1 - (x - 1)2 ] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\)
=> \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy x = {1; 2; 0}
a,2^12.3^54^6.81/(2^2.3)^6+8^4.3^5
b, 4.(-1/2)^3-2.(-1/2)^2+3.(-1/2)+1
c, 5x-7=3x+9
d, 5x-|9-7x|=3
e, -5+|3x-1|+6=|-4|
g, (x-1)^2=(x-1)^4
h, 5^-1.25=125
i, |x+1|+|x+2|+|x+3|=4x
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
1)x^6+3x^5+4x^4+4x^3+4x^2+3x+1
2)(x+y+z)^2+(x-2)^2
3)(a-b)^3+(b-c)^3+(c-a)^3
4)10(x^7+y^7+z^7)=7(x^2+y^2+z^2)(x^5+y^5+z^5)
Tìm số nguyên x,y biết:
a)3x+xy-y-5=0
b) A= 5x- 2\x - 3 thuộc Z
c) B = 6x - 1\ 3x + 2 thuộc Z
d) C= 10x\ 5x - 2 thuộc Z
e) D= 19\ x- 1 * x\9 thuộc Z
f) E = 4x + 5 \ x - 3 thuộc Z
b) A=\(\frac{5x-2}{x-3}=\frac{5x-15+13}{x-3}=\frac{5x-15}{x-3}+\frac{13}{x-3}=\frac{5\left(x-3\right)}{x-3}+\frac{13}{x-3}=5+\frac{13}{x-3}\)
Để A thuộc Z thì \(5+\frac{13}{x-3}\in Z\)
=>13 chia hết cho x-3
=>x-3 \(\in\)Ư(13)={-1;1;-13;13}
x-3=-1 x-3=1 x-3 =-13 x-3=13
x =-1+3 x =1+3 x =-13+3 x =13+3
x=2 x =4 x=-10 x=16
Vậy x=2;4;-10;16 thì A thuộc Z
c)B=\(\frac{6x-1}{3x+2}=\frac{6x+4-5}{3x+2}=\frac{6x+4}{3x+2}+\frac{-5}{3x+2}=\frac{2\left(3x+2\right)}{3x+2}+\frac{-5}{3x+2}=2+\frac{-5}{3x+2}\)
Để B thuộc Z thì \(2+\frac{-5}{3x+2}\in Z\)
=>-5 chia hết cho 3x+2
=>3x+2\(\in\)Ư(-5)={-1;1;-5;5}
3x+2=-1 3x+2=1 3x+2=-5 3x+2=5
3x =-3 3x =-1 3x =-7 3x =3
x =-1 x =-1/3 x =-7/3 x =1
Vậy x=-1;-1/3;-7/3;1 thì B thuộc Z
d) C=\(\frac{10x}{5x-2}=\frac{10x-4+4}{5x-2}=\frac{10-4}{5x-2}+\frac{4}{5x-2}=\frac{2\left(5x-2\right)}{5x-2}+\frac{4}{5x-2}=2+\frac{4}{5x-2}\)
Để C thuộc Z thì \(2+\frac{4}{5x-2}\in Z\)
=> 4 chia hết cho 5x-2
=>5x-2\(\in\)Ư(4)={-1;1;-2;2;-4;4}
5x-2=-1 5x-2=1 5x-2=2 5x-2=-2 5x-2=4 5x-2=-4
bạn tự giải tìm x như các bài trên nhé
d) bạn ghi đề mjk ko hjeu
e)E=\(\frac{4x+5}{x-3}=\frac{4x-12+17}{x-3}=\frac{4x-12}{x-3}+\frac{17}{x-3}=\frac{4\left(x-3\right)}{x-3}+\frac{17}{x-3}=4+\frac{17}{x-3}\)
Để E thuộc Z thì\(4+\frac{17}{x-3}\in Z\)
=>17 chia hết cho x-3
=>x-3 \(\in\)Ư(17)={1;-1;17;-17}
x-3=1 x-3=-1 x-3=17 x-3=-17
bạn tự giải tìm x nhé
điều cuối cùng cho mjk ****
tìm x
a) 5x - 7 = 3x + 9
b) \(\left(x+\frac{1}{2}\right)^2\)= \(\frac{4}{25}\)
c) 5x- /9-7x/ = 3
d) -5+/3x-1/+6=/-4/
e) \(\left(x-1\right)^2=\left(x-1\right)^4\)
g) \(5^{-1}.25^x=125\) ( x thộc Z )
i, /x+1/+/x+2/+/x+3/=4x
/ / là giá trị truyệt đối
. là dấu nhân
mong mn giúp đỡ mik đang gấp
a) \(5x-7=3x+9\)
\(\Rightarrow5x-3x=9+7\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=16:2\)
\(\Rightarrow x=8\)
Vậy \(x=8.\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}-\frac{1}{2}\\x=\left(-\frac{2}{5}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}.\)
c) \(5x-\left|9-7x\right|=3\)
\(\Rightarrow\left|9-7x\right|=5x-3\)
\(\Rightarrow\left[{}\begin{matrix}9-7x=5x-3\\9-7x=3-5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9+3=5x+7x\\9-3=-5x+7x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12=12x\\6=2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=12:12\\x=6:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}.\)
d) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Rightarrow-5+\left|3x-1\right|+6=4\)
\(\Rightarrow-5+\left|3x-1\right|=4-6\)
\(\Rightarrow-5+\left|3x-1\right|=-2\)
\(\Rightarrow\left|3x-1\right|=\left(-2\right)+5\)
\(\Rightarrow\left|3x-1\right|=3.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=3\\3x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:3\\x=\left(-2\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
Bài 1.Tìm x biết
a,(x+8)(x+6)=104+x^2
b,(x+1)(x+2)-(x-3)(x+4)=6
c,4x(x-5)-(x-1)(4x-3)=5
d,(3x-4)(x-2)=3x(x-9)-3
e,(x-5)(x-4)-(+1)(x-2)=7
f,5x(x-3)=(x-2)(5x-1)-5
a) \(\left(x+8\right)\left(x+6\right)=104+x^2\Leftrightarrow x^2+6x+8x+48=104+x^2\)
\(\Leftrightarrow x^2+6x+8x-x^2=104-48\Leftrightarrow14x=56\Leftrightarrow x=\dfrac{56}{14}=4\)
vậy \(x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-\left(x^2+4x-3x-12\right)=6\)
\(\Leftrightarrow x^2+2x+x+2-x^2-4x+3x+12=6\)
\(\Leftrightarrow2x+14=6\Leftrightarrow2x=6-14=-8\Leftrightarrow x=\dfrac{-8}{2}=-4\)
vậy \(x=-4\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left(4x^2-3x-4x+3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x-3=5\Leftrightarrow-13x=5+3=8\Leftrightarrow x=\dfrac{8}{-13}=\dfrac{-8}{13}\)
vậy \(x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)
\(\Leftrightarrow3x^2-6x-4x-3x^2+27x=-3-8\)
\(\Leftrightarrow17x=-11\Leftrightarrow x=\dfrac{-11}{17}\) vậy \(x=\dfrac{-11}{17}\)
e) câu này đề bị thiếu rồi nha bn
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-x-10x+2-5\)
\(\Leftrightarrow5x^2-15x-5x^2+x+10x=2-5\)
\(\Leftrightarrow-4x=-3\Leftrightarrow x=\dfrac{-3}{-4}=\dfrac{3}{4}\) vậy \(x=\dfrac{3}{4}\)
a) \(\left(x+8\right)\left(x+6\right)=104+x^2\)
\(\Leftrightarrow x^2+14x+48=104+x^2\)
\(\Leftrightarrow14x=56\)
\(\Rightarrow x=4\)
b) \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+3x+2-x^2-7x+12=6\)
\(\Leftrightarrow-4x=-8\)
\(\Rightarrow x=2\)
c) \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-4x^2+3x+4x-3=5\)
\(\Leftrightarrow-13x=8\)
\(\Rightarrow x=\dfrac{-8}{13}\)
d) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Leftrightarrow17x=-11\)
\(\Rightarrow x=\dfrac{-11}{17}\)
e) \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)
\(\Leftrightarrow-8x=-15\)
\(\Rightarrow x=\dfrac{15}{8}\)
f) \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Leftrightarrow5x^2-15x=5x^2-11x+2-5\)
\(\Leftrightarrow-4x=-3\)
\(\Rightarrow x=\dfrac{3}{4}\)
Bài 1: Thực hiện phép tính
a) (3x-1)(9x2+3x+1)-4x(x-5)
b) (7x+2)(3-4x)-(x+3)(x2-3x+9)
c) (4x+3)(4x-3)-(2-x)(4+2x+x2)
d) (3x-8)(-5x+6)-(4x+1)(3x-2)
e) (3x-6)4x-2x(3x+5)-4x2
f) (5x-6)(6x-5)-x(3x+10)
Bài 2 : Tính
a) x(x+3)-x2=6
b) 2x(x-5)+x(-2x-1)=6
c) x (x+5)-(x+1)(x-2)=7
d)(3x+4)(6x-3)-(2x+1)(9x-2)=10
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2