tìm x
2 2 x : 2 2 = 10 x 32 + 6 x 2 4
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a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)
b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)
\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)
\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)
\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)
Bài 1: Tính
a) =1\(\frac{49}{60}\)
b) =2\(\frac{13}{30}\)
Bài 2: Tìm x
a) =4\(\frac{1}{21}\)
Riêng câu b) thì mk nghĩ là bạn viết lộn vì mk thấy cái chỗ xx3/4 là mk ko hiểu rồi
\(6-2\left(x-1\right)=4\)
\(\Rightarrow2\left(x-1\right)=6-4\)
\(\Rightarrow2\left(x-1\right)=2\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1=2\)
________________
\(2\cdot\left(x-2\right)+1=7\)
\(\Rightarrow2\cdot\left(x-2\right)=7-1\)
\(\Rightarrow2\cdot\left(x-2\right)=6\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=3+2=5\)
_______________
\(\left(2\cdot x-3\right)+4=9\)
\(\Rightarrow2\cdot x-3=5\)
\(\Rightarrow2\cdot x=3+5\)
\(\Rightarrow2\cdot x=8\)
\(\Rightarrow x=\dfrac{8}{2}=4\)
________________
\(\left(3\cdot x-2\right)-1=3\)
\(\Rightarrow3\cdot x-2=3+1\)
\(\Rightarrow3\cdot x-2=4\)
\(\Rightarrow3\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{3}=2\)
a: =>2(x-1)=2
=>x-1=1
=>x=2
b: =>2(x-2)=6
=>x-2=3
=>x=5
c; =>2x-3=5
=>2x=8
=>x=4
d: =>3x-2=4
=>3x=6
=>x=2
e: =>2(6-x)=4
=>6-x=2
=>x=4
f: =>x-2=5
=>x=7
g: =>10-2x=4
=>2x=6
=>x=3
h: =>2x+4=3
=>2x=-1
=>x=-1/2
j: =>x+2=12
=>x=10
l: =>2x+3=3
=>2x=0
=>x=0
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
a) Số số hạng của dãy A là : ( 160 - 2 ) : 2 + 1 = 80 ( Số hạng )
Tổng của dãy A là : ( 160 + 2 ) x 80 : 2 = 6480
Số số hạng của dãy B là : ( 151 - 1 ) : 3 + 1 = 51 ( số hạng )
Tổng của dãy B là : ( 151 + 1 ) x 51 : 2 = 3876
Có C = ( 2 + 4 - 6 ) + ( 8 + 10 - 12 ) + ( 14 + 16 - 18 ) + .... + ( 116 + 118 - 120)
C = 0 + 6 + 12 + .... + 114
Số số hạng của dãy là : ( 114 - 0 ) : 6 + 1 = 20 ( số hạng )
Tổng của dãy C là : ( 114 + 0 ) x 20 : 2 =1140
b) 5x < 32
\(\Rightarrow\) x < 6,4
\(\Rightarrow\)x lớn nhất là 6
32 < 5x
\(\Rightarrow\)x nhỏ nhất là 7
x:5 = x : 7
\(\Rightarrow\)x = 0
\(2^{2x}:2^2=10.32+6.2^4\)
\(\Leftrightarrow\)\(2^{2x-2}=2^5.10+2^5.3\)
\(\Leftrightarrow\)\(2^{2x-2}=2^5\left(10+3\right)\)
\(\Leftrightarrow\)\(2^{2x-2}:2^5=13\)
\(\Leftrightarrow\)\(2^{2x-7}=13\)
Đề nhầm ???
\(2^{2x}:2^2=10.32+6\cdot2^4\)
\(2^{2x-2}=2^5\left(10+3\right)\)
\(2^{2x-2}:2^5=13\)
\(2^{2x-2-5}=13\)
\(2^{2x-7}=13\)
???