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a) Để A lớn nhất thì \(\frac{15}{4.\left|3x+7\right|+3}\) lớn nhất hay 4.|3x + 7| + 3 nhỏ nhất
Có: \(4.\left|3x+7\right|+3\ge3\forall x\)
Dấu "=" xảy ra khi |3x + 7| = 0
=> 3x + 7 = 0
=> 3x = -7
\(\Rightarrow x=\frac{-7}{3}\)
Với x = \(\frac{-7}{3}\) thay vào đề bài ta được A = 10
Vậy \(A_{Max}=10\) khi x = \(\frac{-7}{3}\)
b) Để B lớn nhất thì \(\frac{21}{8.\left|15x-21\right|+7}\) lớn nhất hay 8.|15x - 21| + 7 nhỏ nhất
Có: \(8.\left|15x-21\right|+7\ge7\forall x\)
Dấu "=" xảy ra khi |15x - 21| = 0
=> 15x - 21 = 0
=> 15x = 21
\(\Rightarrow x=\frac{21}{15}=\frac{7}{5}\)
Với \(x=\frac{7}{5}\) thay vảo đề bài ta tìm được B = \(\frac{8}{3}\)
Vậy \(B_{Max}=\frac{8}{3}\) khi x = \(\frac{7}{5}\)
c) Có: \(\begin{cases}\left|x+1\right|\ge x+1\\\left|3x-4\right|\ge4-3x\\\left|2x-1\right|\ge2x-1\end{cases}\)\(\forall x\)
\(\Rightarrow C\ge\left(x+1\right)+\left(4-3x\right)+\left(2x-1\right)+5\)
hay \(C\ge9\)
Dấu "=" xảy ra khi \(\begin{cases}x+1\ge0\\3x-4\le0\\2x-1\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\3x\le4\\2x\ge1\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\x\le\frac{3}{4}\\x\ge\frac{1}{2}\end{cases}\)\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{4}\)
Vậy \(C_{Max}=9\) khi \(\frac{1}{2}\le x\le\frac{3}{4}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a) \(\frac{4}{x+5}=\frac{3}{2x-1}\)
=> 4(2x - 1) = 3(x + 5)
=> 8x - 4 = 3x + 15
=> 8x - 3x = 15 + 4
=> 5x = 19
=> x = 19/5
b) \(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)
=> \(\left(\frac{x+11}{19}-1\right)+\left(\frac{x+12}{20}-1\right)+\left(\frac{x+13}{21}-1\right)=0\)
=> \(\frac{x-8}{19}+\frac{x-8}{20}+\frac{x-8}{21}=0\)
=> \(\left(x-8\right)\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
=> x - 8 = 0
=> x = 8
c) \(\left(2x-1\right)^2=\left(2x-1\right)^3\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^3=0\)
=> \(\left(2x-1\right)^2.\left[1-\left(2x-1\right)\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\1-2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
a) 4/x + 3 = 3/2x - 1
<=> 4.(2x - 1) = (x + 3).3
<=> 8x - 4 = 3x + 9
<=> 8x = 3x + 9 + 4
<=> 8x = 3x + 13
<=> 8x - 3x = 13
<=> 5x = 13
<=> x = 13/5
=> x = 13/5
c) (2x - 1)2 = (2x - 1)3
<=> 4x2 - 4x + 1 = 8x3 - 12x2 + 6x - 1
<=> 8x3 - 12x2 + 6x - 1 = 4x2 - 4x + 1
<=> 8x3 - 12x2 + 6x - 1 - 1 = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x - 4x = 4x2
<=> 8x3 - 12x2 + 10x - 2 = 4x2
<=> 8x3 - 12x2 + 10x - 2 - 4x2 = 0
<=> 8x2 - 16x2 + 10x - 2 = 0
<=> 2(x - 1)(2x - 1)2 = 0
<=> x - 1 = 0 hoặc 2x - 1 = 0
x = 0 + 1 2x = 0 + 1
x = 1 2x = 1
x = 1/2
=> x = 1 hoặc x = 1/2
\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)
\(\Leftrightarrow7-6x=3x-5\)
\(\Leftrightarrow7+5=3x+6x\)
\(\Leftrightarrow12=9x\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)