Cau 1 tim x biet
a) |2x-2019|=1
b)(2-x)5=-32
c) 4x+22x+1-22x+2=-1024
tìm x
5)
4x x 5 x 4x 3 5
6)
2
2
x 2 x 1 6
7)
2
3
(3 2) 3 .
4
x x x
8) (3x + 1). (2x- 3) – 6x.(x + 2) = 16
8: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
tim x
-2x(6x-2)+3x(4x-7)=8
(7x-2)(2x+3)-(3x-5)(4x+6)=2x2-5
+(-2x+3)-6x(3x-4)=-22x2+7x
a)(6x^2+17x+12):(2x+3) b)(5x^2+13x-6):(5x-2) c)(-8x^2+22x-15):(2x-5) d)(14x^2-33x-5):(2x-5) e)(2x^3+7x^2+15x+6):(2x+1) f)(x^3+4x^2-11x-2):(x-2) g)(12x^3+2x^2+4x+3):(2x+1)
a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)
=3x+4
b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)
\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)
c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)
d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)
=7x+1
e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)
\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)
f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)
g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)
cíu tui zới
tui hứa sẽ tick
3) 2x + 3 + 2x = 36
4) 4x+1 - 22x = 12
5) 5x+3 - 5x+1= 3000
4) 4x+1 - 22x
3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
Phương trình 4 x - 2 x + 1 + 2 2 x - 1 sin 2 x + y - 1 + 2 = 0 có nghiệm x = a y = b . Tính S = a + b
A. S = π 2 + k π
B. S = - π 2 + k 2 π
C. S = π 3 + k π
D. S = - π 3 + k 2 π
pt da thuc sau thanh nhan tu : x^3-9x^2+15x+ 25 ; X^8 - 4X^2- 11X+ 30
2 X^4+X^3 - 22X^2+ 15 X- 36
3X^3+ 5X^2-14X+ 4
2X^3- X^2- 3X -1
MN GIUPS MINH CAU NAY NHA ,MINH DANG CAN GAP, CHIEU NAY NOP
a) \(x^3-9x^2+15x+25\)
\(=x^3+x^2-10x^2-10x+25x+25\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2.x.5+25\right)=\left(x+1\right)\left(x-5\right)^2\)
b) \(x^8-4x^2-11x+30:\text{đề sai thì phải bạn ạ!}\)
c) \(x^4+x^3-22x^2+15x-36\)
\(=\left(2x^4-6x^3\right)+\left(7x^3-21x^2\right)-\left(x^2-3x\right)+12x-36\)
\(=2x^3\left(x-3\right)+7x^2\left(x-3\right)-x\left(x-3\right)+12\left(x-3\right)\)
\(=\left(x-3\right)\left(2x^3+8x^2-x^2-x+12\right)\)
\(=\left(x-3\right)\left[\left(2x^3+8x^2\right)-\left(x^2+4x\right)+3\left(x+4\right)\right]\)
\(=\left(x-3\right)\left[2x^2\left(x+4\right)-x\left(x+4\right)+3\left(x+4\right)\right]\)
\(=\left(x-3\right)\left(x+4\right)\left(2x^2-x+3\right)\)
giải pt
a) \(sin^2x+2sin^22x+sin^23x-2=0\)
b) \(2cosx.cos\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
c) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)
d) \(1+cot2x=\frac{1-cos2x}{sin^22x}\)
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)
\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)
\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
quy đồng mẫu thức phân thức
2/x^2-5x+6 và 3/x-3
x^2-4x+4/x^2-2x và x+1/x^2-1
x^3-2^3/x2-4 và 3/x+2
2x/x2+3x+2 và 3x/x2+4x+3
giải các pt
a) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)
b) \(\sqrt{3}tanx+cotx-\sqrt{3}-1=0\)
c) \(6sin^2x+2sin^22x=5\)
d) \(cos^22x+cos^2\left(x-\frac{\pi}{4}\right)-1=0\)
e) \(\left(1+tan^2x\right)\left(9-13cosx\right)+4=0\)
a/
\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)
\(\Leftrightarrow3+5cosx=\left(1-cos^2x\right)-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0\)
\(\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0\)
\(a+b+c=\sqrt{3}-\left(\sqrt{3}+1\right)+1=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow6\left(\frac{1-cos2x}{2}\right)+2\left(1-cos^22x\right)=5\)
\(\Leftrightarrow-2cos^22x-3cos2x=0\)
\(\Leftrightarrow cos2x\left(2cos2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
d/
\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)
\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^22x+sin2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)