Tìm x biết
2x+2x+1+2x+2=896
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>2x^3=58-4=54
=>x^3=27
=>x=3
b; =>(5-x)^5=2^5
=>5-x=2
=>x=3
c: =>(5x-6)^3=4^3
=>5x-6=4
=>5x=10
=>x=2
d: (3x)^3=(2x+1)^3
=>3x=2x+1
=>x=1
1=>2x3=54
=>x3=27 =>x=3
2=>(5-x)5=25
=>5-x=2
=>x=3
3=>(5x-6)3=43
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1
\(2x-10,01=19,01-3\\ \Rightarrow2x-10,01=16,01\\ \Rightarrow2x=16,01+10,01\\ \Leftrightarrow2x=26,02\\ \Leftrightarrow x=26,02:2=13,01\)
2x-10,01=19,01-3
=> 2x-10,01= 16,01
=> 2x= 16,01+10,01
=>2x= 26,02
=> x= 26,02: 2= 13,01
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
==========
\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)
\(\Leftrightarrow x=25\)
b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Leftrightarrow-9x=1\)
hay \(x=-\dfrac{1}{9}\)
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x^2+3x+1-2x^2-x+3=0\)
=>2x=-4
hay x=-2
\(2^x+2^{x+1}+2^{x+2}=896\)
\(\Rightarrow2^x\times1+2^x\times2+2^x\times2^2=896\)
\(\Rightarrow2^x\left(1+2+2^2\right)=896\)
\(\Rightarrow2^x\times7=896\)
\(\Rightarrow2^x=\frac{896}{7}\)
\(\Rightarrow2^x=128\)
\(\Rightarrow2^x=2^7\)
\(\Rightarrow x=7\)