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\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)
Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
\(x=0\)
Lập bảng xét dấu :
\(x\) \(0\) \(\dfrac{9}{16}\)
\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\) \(-\) \(0\) \(-\) \(0\) \(+\)
\(\left|x\right|\) \(-\) \(0\) \(+\) \(0\) \(+\)
TH1 : \(x< 0\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))
TH2 : \(0\le x\le\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))
TH3 : \(x>\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))
Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)
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A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
x | \(\frac{-2}{3}\) 3 |
3x+2 | - 0 + | + |
x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Câu hỏi của Jey - Toán lớp 7 - Học toán với OnlineMath
2/ \(\left(a-b\right)^2+6ab=36\Rightarrow6ab=36-\left(a-b\right)^2\le36\Rightarrow ab\le\frac{36}{6}=6\)
Dấu "=" xảy ra khi \(\orbr{\begin{cases}a=b=\sqrt{6}\\a=b=-\sqrt{6}\end{cases}}\)
Vậy abmax = 6 khi \(\orbr{\begin{cases}a=b=\sqrt{6}\\a=b=-\sqrt{6}\end{cases}}\)
3/
a, Để A đạt gtln <=> 17/13-x đạt gtln <=> 13-x đạt gtnn và 13-x > 0
=> 13-x = 1 => x = 12
Khi đó \(A=\frac{17}{13-12}=17\)
Vậy Amax = 17 khi x = 12
b, \(B=\frac{32-2x}{11-x}=\frac{22-2x+10}{11-x}=\frac{2\left(11-x\right)+10}{11-x}=2+\frac{10}{11-x}\)
Để B đạt gtln <=> \(\frac{10}{11-x}\) đạt gtln <=> 11-x đạt gtnn và 11-x > 0
=>11-x=1 => x=10
Khi đó \(B=\frac{10}{11-10}=10\)
Vậy Bmax = 10 khi x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
=2x^3-2x^2-5x-10-2x^2+4x+x^2(2x-3)-x(x+1)-3x+2
=2x^3-4x^2-4x-8+2x^3-6x^2-x^2+x
=4x^3-11x^2-3x-8
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
![](https://rs.olm.vn/images/avt/0.png?1311)
=>|3x-2/5|=1/35+90/35=91/35
=>3x-2/5=91/35 hoặc 3x-2/5=-91/35
=>3x-2/5=13/5 hoặc 3x-2/5=-13/5
=>3x=15/5=3 hoặc 3x=-11/5
=>x=-11/5 hoặc x=1
Lời giải:
$|3x-\frac{2}{5}|=\frac{1}{35}+\frac{18}{7}=\frac{13}{5}$
$\Rightarrow 3x-\frac{2}{5}=\frac{13}{5}$ hoặc $3x-\frac{2}{5}=\frac{-13}{5}$
$\Rightarrow 3x=3$ hoặc $3x=\frac{-11}{5}$
$\Rightarrow x=1$ hoặc $x=\frac{-11}{15}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2\left|3x-1\right|+1=5\)
\(\Leftrightarrow2\left|3x-1\right|=4\)
\(\Leftrightarrow\left|3x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
2.l3x-1l+1=5
=>2.l3x-1l=4
=>l3x-1l=2
TH1:3x-1=2 =>x=1
TH2:3x-1=-2 =>x=-1/3