a) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\\ \)
b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)
c) \(\left(0,4x-1,3\right)^2=5,29\)
Tìm x,y biết :
a)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
b)\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
c)\(\left(0,4x-1,3\right)^2=5,29\)
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
TH1: \(0,4x-1,3=2,3\)
\(0,4x=3,6\)
\(x=9\)
TH2: \(0,4x-1,3=-2,3\)
\(0,4x=-1\)
\(x=-\dfrac{5}{2}\)
=.= hok tốt!!
BT3: Tìm x
\(a,\left(x+2\right)^2-9=0\)
\(b,x^2-2x+1=25\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
a) \(2sin\left(x+\dfrac{\pi}{3}\right)+1=0\)
b) \(1+2sin\left(x-30^o\right)=0\)
c) \(\sqrt{3}+2sin\left(x-\dfrac{\pi}{6}\right)=0\)
d) \(2sin\left(x+10^o\right)+\sqrt{3}=0\)
e) \(\sqrt{2}+2sin\left(x-15^o\right)=0\)
f) \(\sqrt{2}sin\left(x-\dfrac{\pi}{3}\right)+1=0\)
g) \(3+\sqrt{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)
h) \(1+sin\left(x-30^o\right)=0\)
i) \(3+\sqrt{5}sin\left(x-\dfrac{\pi}{6}\right)=0\)
k) \(2\sqrt{2}sin^2x-sin2x=0\)
a: =>2sin(x+pi/3)=-1
=>sin(x+pi/3)=-1/2
=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi
=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi
b: =>2sin(x-30 độ)=-1
=>sin(x-30 độ)=-1/2
=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ
=>x=k*360 độ hoặc x=240 độ+k*360 độ
c: =>2sin(x-pi/6)=-căn 3
=>sin(x-pi/6)=-căn 3/2
=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi
=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi
d: =>2sin(x+10 độ)=-căn 3
=>sin(x+10 độ)=-căn 3/2
=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ
=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ
e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)
=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)
=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ
=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ
f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)
=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi
=>x=pi/12+k2pi hoặc x=19/12pi+k2pi
g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
h) \(1+sin\left(x-30^o\right)=0\)
\(\Leftrightarrow sin\left(x-30^o\right)=-1\)
\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow x=-60^0+k360^o\)
tìm GTLN
a)\(A=x^2+5y^2+2xy-4x-8y+2015\)
b)\(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
c)\(C=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)
d)\(D=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)
\(B=2x^2+2x+2012^2+2013^2\)
\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)
\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)
\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)
Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
TÌM X,BIẾT:
a/\(\left(5x+1^{ }\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
b/\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
1) \(3^x+3^{x+2}=810\)
2) \(\left(\dfrac{2}{3}x-1\right).\left(0,4x+5\right)=0\)
\(1,\Leftrightarrow3^x\left(1+3^2\right)=810\\ \Leftrightarrow3^x=\dfrac{810}{10}=81=3^4\\ \Leftrightarrow x=4\\ 2,\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=1\\0,4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{25}{2}\end{matrix}\right.\)
Tìm x, biết:
a) \(\left|x-24\right|+\left|y+8\right|=1\)
b)\(\left(x-2\right)^{10}+\left|y-2\right|=0\)
c)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
d)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+2017+2018=2018\)
Giải thích cụ thể giúp mk nha
Rút gọn rồi tính giá trị biểu thức :
a) \(A=\left(x+3\right)^2+\left(x-3\right).\left(x+3\right)-2.\left(x+2\right).\left(x-4\right)\); với x = \(-\frac{1}{2}\)
b) \(B=\left(3x+4\right)^2-\left(x-4\right).\left(x+4\right)-10x\); với x = \(-\frac{1}{10}\)
c) \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3.\left(x-2\right).\left(x+2\right)\); với x = 1
d) \(D=\left(x-3\right).\left(x+3\right)+\left(x-2\right)^2-2x.\left(x-4\right)\); với x = -1