C= /2x-1/-/2x-15/
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a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)
\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)
\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)
\(\Rightarrow x=\frac{23}{20}\)
b) \(x+\frac{46}{15}=1,5\)
\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)
\(\Rightarrow x=\frac{-47}{30}\)
c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)
\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)
\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)
a) 2x . 4 = 128
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b) x15 = x1
<=> x15 - x = 0
<=> x(x14 - 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
c) (2x + 1)3 = 125
<=> (2x + 1)3 = 53
<=> 2x + 1 = 5
<=> 2x = 4
<=> x = 2
d) (x - 5)4 = (x - 5)6
<=> (x - 5)6 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)2 - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5
Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{5;-2\right\}\)
\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)
Câu d xem lại đề
\(C=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{\left(2x+1\right)\times\left(2x+3\right)}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{\left(2x+1\right)\times\left(2x+3\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow x=45\).
`@` `\text {Ans}`
`\downarrow`
`a,`
`x^2 + 2x + 1 = 9`
`=> x^2 + 2x + 1 - 9 = 0`
`=> x^2 + 2x - 8 = 0`
`=> x^2 + 4x - 2x - 8 = 0`
`=> (x^2 + 4x) - (2x + 8) = 0`
`=> x(x + 4) - 2(x + 4) = 0`
`=> (x-2)(x+4) = 0`
`=>`\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy, `x \in {2; 4}`
`b,`
`x^2 - 1 = 15`
`=> x^2 = 15 + 1`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = +-4`
Vậy, `x \in {4; -4}`
`c)`
`19 - 2x^2 = 1`
`=> 2x^2 = 19 - 1`
`=> 2x^2 = 18`
`=> x^2 = 18 \div 2`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {3; -3}.`
a) \(x^{10}=x\)
\(\Rightarrow x=1;0\)
b) \(x^{10}=1^x\)
\(\Rightarrow x^{10}=1\)
\(\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
Vì \(\left(2x-15\right)^5=\left(2x-15\right)^3\) nên kết quả của ( 2x - 15 ) chỉ có thể = 0 hoặc 1
Nếu (2x-15) = 0 thì thay vào đó ta được :
\(2x-15=0\)
\(2x=15\)
\(\Rightarrow x=15:2=7\left(dư1\right)\)loại
nếu kết quả ( 2x-15) = 1 thì thay vào đó ta được :
\(2x-15=1\)
\(2x=1+15\)
\(2x=16\)
\(\Rightarrow x=16:2=8\)
Vậy x = 8
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`