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a: \(\Leftrightarrow\left|x-3\right|=12-5x-8=-5x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(-5x+4\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
b: \(\left(\sqrt{x}+3\right)^{10}=1024\cdot125^2\cdot25^2\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^{10}=2^{10}\cdot5^6\cdot5^4=10^{10}\)
\(\Leftrightarrow\sqrt{x}+3=10\)
hay x=49
c: \(\dfrac{3-0.2x}{5}=\dfrac{7}{15}+1.4x\)
\(\Leftrightarrow\dfrac{9-0.6x}{15}=\dfrac{7}{15}+\dfrac{21x}{15}\)
=>21x+7=9-0,6x
=>21,6x=-2
hay x=-5/54
d: \(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{5^9\cdot7^9\left(4\cdot7-5^2\right)}{5^9\cdot7^9\cdot4}\)
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{28-25}{4}=\dfrac{3}{4}\)
=>3x=-1
hay x=-1/3
=> 6x - 3 - 5 - 15x = 44
=> -9x - 8 = 44
=> -9x = 52
=> x = \(\frac{-52}{9}\)
nhớ
3(2x-1)-5(1+3x)=44
\(\Leftrightarrow\)6x-3-5-15x=44
\(\Leftrightarrow\)-11x=52
\(\Leftrightarrow\)x=-52/11
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
\(VT=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\) \(\Rightarrow VT\ge4\) (1)
Lại có \(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{8}{3\left(x+1\right)^2+2}\le\dfrac{8}{2}=4\) \(\Rightarrow VP\le4\) (2)
Từ (1), (2) \(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\left|2x+3\right|+\left|2x-1\right|=4\\\dfrac{8}{3\left(x+1\right)^2+2}=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(1-2x\right)\ge0\\3\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
Vậy pt có nghiệm duy nhất \(x=-1\)
Ta có: \(\left|2x-1\right|-x=4\)
\(\Rightarrow\left|2x-1\right|=4+x\)
+) TH1: \(2x-1\ge0\Rightarrow2x\ge1\Rightarrow x\ge\dfrac{1}{2}\)
Ta có: \(2x-1=4+x\)
\(\Rightarrow2x-x=1+4\)
\(\Rightarrow x=5\) (t/m)
+) TH2: \(2x-1< 0\Rightarrow2x< 1\Rightarrow x< \dfrac{1}{2}\)
Khi đó \(-2x+1=4+x\)
\(\Rightarrow-2x-x=-1+4\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-1\) (t/m)
Vậy \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\).
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
a, Xét : x-4 = 0 => x= 4
2x+1 = 0 => x= \(\frac{1}{2}\)
x+3 = 0 => x = -3
x + 9 = 0 => x = -9
Khi đó ta có bảng xét dấu :
| x | -9 | -3 | \(\frac{1}{2}\) | 4 |
| x-4 | -13 | -7 | \(\frac{-7}{2}\) | 0 |
| 2x+1 | -17 | -5 | 2 | 9 |
| x+3 | -6 | 0 | \(\frac{7}{2}\) | 7 |
| x+9 | 0 | 6 | \(\frac{19}{2}\) | 13 |
=> có 5 trường hợp:
TH1 : \(x\le-9\)
TH2 : \(-9\le x< -3\)
TH3 : \(-3\le x< \frac{1}{2}\)
TH4 : \(\frac{1}{2}\le x< 4\)
Do đó :
TH1 : \(x\le-9\)
Ta có : /x-4/ = -(x-4) = 4 - x
/2x+1/ = -(2x+1) = -2x -1
/x+3/ = -(x + 3 ) = -x - 3
/x-9/ = -(x-9) = -x + 9 Thay vào đề bài ta có:
3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5
=> 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5
=>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5
=> -13 - 7x = 5
7x = -13 - 5
7x = -18
x = \(\frac{-18}{7}\)( Ko TM)
Tương tự với 4 trường hợp còn lại.
a)TH1 x>=3 \(\left|x-3\right|\)=x-3
pttt: x-3-2x=1 suy ra x=-4 <3 -> loại
TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn
b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)
VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)
vậy pt đã cho tương đương với 4^x=9^x \(\Leftrightarrow\left(\dfrac{4}{9}\right)\)^x =1 suy ra x =0
bạn ơi: pttt, vt, vp là j vậy???