tim gtln 5+\(\dfrac{15}{4\left|3x+7\right|+3}\)
Những câu hỏi liên quan
tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
1. dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
2. x-dfrac{3left(x+30right)}{15}-24dfrac{1}{2}dfrac{7x}{10}-dfrac{2left(10x+2right)}{5}
3. 14dfrac{1}{2}-dfrac{2left(x+3right)}{5}dfrac{3x}{2}-dfrac{2left(x-7right)}{3}
4. dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
5. dfrac{3left(2x-1right)}{4}-dfrac{3x+1}{10}+1dfrac{2left(3x+2right)}{5}
6. x-dfrac{3}{17}left(2x-1right)dfrac{7}{34}left(1-2xright)+dfrac{10x-3}{2}
7. dfrac{3le...
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1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
\(a,\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
b,\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)
c,\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)
a:
\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)
\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)
\(\Leftrightarrow0x=88\)
Vậy pt vô nghiệm
b:
\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)
\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)
\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)
\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)
\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)
\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)
Vậy tập nghiện của pt S= {-38}
c:
\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)
\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)
\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)
\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)
\(\Leftrightarrow0x=0\)
Vậy pt vô số nghiệm
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giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
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\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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1. 25dfrac{1}{7}:left(-dfrac{5}{7}right)-15dfrac{1}{7}:left(-dfrac{5}{7}right)+dfrac{4}{5} 3. 2dfrac{2}{3}:left{left[left(3,72-0.02right)dfrac{10}{37}right]:dfrac{5}{6}+2,8right}-dfrac{7}{15}2. left(3+dfrac{4}{5}-dfrac{5}{12}right)left(dfrac{6}{7}-dfrac{3}{5}right)^24.23+3.left(-dfrac{1}{2}right)^2-22.4+left[left(-2right)^2:dfrac{1}{2}right]
Đọc tiếp
1. \(25\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)-15\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)+\dfrac{4}{5}\) 3. \(2\dfrac{2}{3}:\left\{\left[\left(3,72-0.02\right)\dfrac{10}{37}\right]:\dfrac{5}{6}+2,8\right\}-\dfrac{7}{15}\)
2. \(\left(3+\dfrac{4}{5}-\dfrac{5}{12}\right)\left(\dfrac{6}{7}-\dfrac{3}{5}\right)^2\)
4.23+3.\(\left(-\dfrac{1}{2}\right)^2\)-22.4+\(\left[\left(-2\right)^2:\dfrac{1}{2}\right]\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
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Tìm GTLN CỦA BIỂU THỨC 5+\(\frac{15}{4\left|3x+7\right|+3}\)
Ta có: \(\left|3x+7\right|\ge0\forall x\)
\(\Rightarrow4\cdot\left|3x+7\right|\ge0\forall x\)
\(\Rightarrow4\cdot\left|3x+7\right|+3\ge3\forall x\)
\(\Rightarrow\frac{15}{4\left|3x+7\right|+3}\le\frac{15}{3}=5\forall x\)
\(\Rightarrow\frac{15}{4\left|3x+7\right|+3}+3\le8\forall x\)
Dấu '=' xảy ra khi \(3x+7=0\)
\(\Leftrightarrow3x=-7\)
hay \(x=-\frac{7}{3}\)
Vậy: GTLN của biểu thức \(\frac{15}{4\left|3x+7\right|+3}\) là 8 khi \(x=-\frac{7}{3}\)
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Giải hệ phương trình
left{{}begin{matrix}4left(2x-y+3right)-3left(x-2y+3right)483left(3x-4y+3right)+4left(4x-2y-9right)48end{matrix}right.
left{{}begin{matrix}6left(x+yright)8+2x-3y5left(y-xright)5+3x+2yend{matrix}right.
left{{}begin{matrix}-2left(2x+1right)+1,53left(y-2right)-6x11,5-4left(3-xright)2y-left(5-xright)end{matrix}right.
left{{}begin{matrix}dfrac{8x-5y-3}{7}+dfrac{11y-4x-7}{5}12dfrac{9x+4y-13}{5}-dfrac{3left(x-2right)}{4}15end{matrix}right.
left{{}begin{matrix}2sqrt{3}x-sqrt{5}y2...
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Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
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