Bài 2: Giải pt, bpt sau:
4) \(8^x \leq 4.(4-2^x)\)
5) \(\left(\dfrac{1}{3}\right)^{\frac{2}{x}} + 3.\left(\dfrac{1}{3}\right)^{\frac{1}{x}+1} > 12\)
1.giải các bpt sau
a.\(\left(x-3\right)\left(x+3\right)\ge x^2-7x+1\)
b.\(\dfrac{1,5-x}{5}\ge\dfrac{4x+5}{2}\)
2.giải các pt sau
\(x^3+1=x.\left(x+1\right)\)
Giải bpt sau
a, \(\left(x+3\right)^2-\left(x-3\right)^2\le3\left(x+1
\right)\)
b, \(2\left(x+3\right).\left(x+4\right)>\left(x-2\right)^2+\left(x-1\right)^2\)
c, \(5x^2-18x+19-\left(2x-3\right)^2>0\)
d, \(\dfrac{\left(3x-2\right)^2}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{-15x\left(5-3x\right)}{2}\)
e, \(2x^2+2x+2-\dfrac{15\left(x-1\right)}{2}-1>2x\left(x-2,75\right)\)
g, \(\dfrac{5x^2-3}{5}+\dfrac{3x-1}{4}< \dfrac{x\left(2x+3\right)}{2}-5\)
giải pt
1.(x^2-x+1)(X^2-x+2)=2
2.X(x+2)(x+3)(x+5)=280
3.(x+3)(x+4)(X+5)=x
4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\) 5. 12/x^2-12/x^2+2=1 6.(x^2-6x)^2+14(x-3)^2=81 7.(x^2+5x)^2-2(x^2+5x)-24=0
8. x^2+2x+3=(x^2+x+1)(X^4+x^2+4)
2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)
\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x+3=t\)
\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)
\(\Leftrightarrow t^2-9=280\)
\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0
\(\Leftrightarrow\) x = 2 hoặc x = - 7
Vậy x = 2 hoặc x = -7.
3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)
\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)
\(\Leftrightarrow x^3+12x^2+46x+60=0\)
\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)
\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)
\(\Leftrightarrow x=-6\)
Vậy x = -6.
4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)
\(\Leftrightarrow2\left[\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right]=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{6}{x\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2x\left(x+6\right)=54\)
\(\Leftrightarrow2x^2+12x-54=0\)
\(\Leftrightarrow2x^2-6x+18x-54=0\)
\(\Leftrightarrow2x\left(x-3\right)+18\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+18\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x+9\right)=0\)
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 9 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -9
Vậy x = 3 hoặc x = -9.
Giải các pt sau:
a, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
b,\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
Giúp mình với ạ
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
Giải các pt và Bpt
a) |x+5|=3x+1
b) \(\dfrac{3\left(x-1\right)}{4}+1\ge\dfrac{x+2}{3}\)
c) \(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
a ) \(\left|x+5\right|=3x+1\) ( 1 )
+ ) \(x+5=x+5.\) Khi \(x\ge-5\)
\(\left(1\right)\Leftrightarrow x+5=3x+1\)
\(\Leftrightarrow-2x=-4\Leftrightarrow x=2\) ( TM )
+ ) \(x+5=-x-5.\) Khi \(x< -5\)
\(\left(1\right)\Leftrightarrow-x-5=3x+1\)
\(\Leftrightarrow-4x=6\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)( KTM )
Vậy ..........
b ) \(\dfrac{3\left(x-1\right)}{4}+1\ge\dfrac{x+2}{3}\)
\(\Leftrightarrow9x-9+12\ge4x+8\)
\(\Leftrightarrow5x\ge5\)
\(\Leftrightarrow x\ge1\)
Vậy ...........
c ) \(\dfrac{x-2}{x+1}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(1\right)\)
ĐKXĐ : \(x\ne2;x\ne-2.\)
\(\left(1\right)\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x-2\right)^2-3\left(x+2\right)=2x-22\)
\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(TMĐKXĐ\right)\)
Vậy .........
\(\Leftrightarrow\)
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
Giải pt sau
\(\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{2x+4}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
ĐKXĐ: ...
\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)
\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các pt sau:
a)\(x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12\)
b) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=5\left(\dfrac{x}{3}+\dfrac{4}{x}\right)\)
c) \(\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}\)
d) \(\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}\)
e) \(x^2+\left(\dfrac{x}{x-1}\right)^2=8\)
g) \(x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)\)
f) \(\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12=0\)
giải pt sau \(\left(\dfrac{x+1}{x-2}\right)^2-3\left(\dfrac{2x-4}{x-4}\right)^2+\dfrac{x+1}{x-4}=0\)
ĐKXĐ: \(x\ne\left\{2;4\right\}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)
Phương trình trở thành:
\(a^2-12b^2+ab=0\)
\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)
\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)
Bạn tự quy đồng và hoàn thành phần còn lại nhé
Giải pt: \(\dfrac{3}{5x-1}+\dfrac{2}{3-5x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\dfrac{5+96}{x^2-16}=\dfrac{2x—1}{x+4}-\dfrac{3x-1}{4-x}\)
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)