Tim x\(3< \left|x-1\right|< 5\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)
\(\Leftrightarrow40x=46+9+25=80\)
\(\Leftrightarrow x=2\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)
\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)
\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)
\(\Leftrightarrow2x=-2+6-5=-1\)
\(\Leftrightarrow x=\frac{1}{2}\)
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ( x + 7 ) + ( x + 9 ) + ( x + 11 ) + (x + 13 ) = 119
x X 7 + ( ! + 3 + 5 + 7 + 9 + 11 + 13 ) = 119
x X 7 + 49 = 119
x X 7 = 119 - 49
x X 7 = 70
x = 70 : 7
x = 10
tk mk và gửi kết bạn cho mk nha huyền trang
các bạn cũng tk và gửi kết bạn cho mk nha
(x+1)+(x+3)+..+(x+13)=119
=>7x+(1+3+..+13) =119
=>7x+49 =119
=>7x =119-49
=>7x =70
=> x =10
Vậy x=10
Chúc bn hok giỏi!
3.(x-1/2) -5(x+3/5)=-x+1/5
3x - 3/2 -5x +3 = -x+1/5
3x-5x+x= 3/2-3+1/5
x.(3-5+1)=15/10 + (-30/10)+2/10
x.(-1)= -13/10
x = -13/10 : (-1)
x=13/10
vậy x=13/10
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
a: \(\Leftrightarrow\left|2x+3\right|-4\left|x-4\right|=5\)
TH1: x<-3/2
Pt sẽ là -2x-3-4(4-x)=5
=>-2x-3-16+4x=5
=>2x-19=5
=>2x=24
hay x=12(loại)
TH2: -3/2<=x<4
Pt sẽ là 2x+3-2(4-x)=5
=>2x+3-8+2x=5
=>4x-5=5
hay x=5/2(nhận)
TH3: x>=4
Pt sẽ là 2x+3-2(x-4)=5
=>2x+3-2x+8=5
=>11=5(loại)
b: TH1: x<-3
Pt sẽ là 1-x-3-x=4
=>-2x-2=4
=>-2x=6
hay x=-3(loại)
TH2: -3<=x<1
Pt sẽ là x+3+1-x=4
=>4=4(luôn đúng)
TH3: x>=1
Pt sẽ là x-1+x+3=4
=>2x+2=4
hay x=1(nhận)
a, 3x+1-2=32+[52-3(22-1)]
3x+1-2=9+[25-3(4-1)]
3x+1-2=9+[25-3.3]
3x+1-2=9+[25-9]
3x+1-2=9+16
3x+1-2=25
3x+1=25+2
3x+1=27
3x+1=33
x+1=3
x=3-1
x=2
b,(32)2+2x=5(5+22.3)
81+2x=5(5+4.3)
81+2x=5(5+12)
81+2x=5.17
81+2x=85
2x=85-81
2x=4
2x=22
x=2
\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu
\(\left(x-\frac{1}{3}\right)\left(y-\frac{1}{2}\right)\left(z-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{2}\\z=5\end{cases}}\)
Vì \(z+3=y+1\Rightarrow y=7\)
Lại có \(y+1=x+2\Rightarrow x=8-2=6\)
Vậy x = 6 ; y = 7 ; z = 5
\(VP=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}=\frac{2\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{2\left(a+5\right)}{4\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+5\right)+2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left[\left(a+3\right)+\left(a+5\right)\right]}{4\left(a+3\right)\left(a+5\right)}=\frac{\left(a+3\right)+\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{\left(a+a\right)+\left(3+5\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{2a+8}{2\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+4\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(VT=\frac{x-2}{\left(a+3\right)\left(a-5\right)}\)
\(\Rightarrow\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(\Rightarrow\frac{x-2}{a+4}=\frac{\left(a+3\right)\left(a-5\right)}{\left(a+3\right)\left(a+5\right)}\Rightarrow\frac{x-2}{a+4}=\frac{a-5}{a+5}\Rightarrow\left(x-2\right)\left(a+5\right)=\left(a-5\right)\left(a+4\right)\)
chịu
Ta có : 3 < | x - 1| < 5
\(\Rightarrow\) |x - 1| = 4
\(\Rightarrow\) x - 1 = 4 \(\Rightarrow\) x = 5
x - 1 = - 4 \(\Rightarrow\) x = - 3
Vậy x \(\in\){5; -3}
\(3< \left|x-1\right|< 5\)
\(\Leftrightarrow\orbr{\begin{cases}3< x-1< 5\\-3>x-1>-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4< x< 6\\-2>x>-4\end{cases}}\)