tìm x
a) 3.2x+(-1.2)x+2.7= -4.9
b)\(\dfrac{37-x}{x+13}\)=\(\dfrac{3}{7}\)
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\(3.2\times x\left(-1.2\right)\times x+2.7=-4.9\) câu a nhé
\(-5.6\times x+2.9\times x-3.86=-9.8\) câu b nhé
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3.2x+(-1.2)*x+2.7=-4.9
(3.2+(-1.2)*x=-4.9-2.7
2*x=-76
x=-76/2
x=-38
3.2.x+1.2.x+2.7=-4.9
6.x+2.x+14=-36
(6+2).x+14=-36
8.x+14=-36
8x=-36-14
8x=-50
x=-50:8
x=-6,25
a: \(x^2-2x+3\)
\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)
=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
b: \(x^2-5x+10\)
\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)
\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)
=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)
=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)
Dấu '=' xảy ra khi \(x-\frac52=0\)
=>\(x=\frac52\)
c: \(x^2-x+6\)
\(=x^2-x+\frac14+\frac{23}{4}\)
\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)
=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)
=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)
Dấu '=' xảy ra khi \(x-\frac12=0\)
=>\(x=\frac12\)
d: \(x^2+x+5\)
\(=x^2+x+\frac14+\frac{19}{4}\)
\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)
Dấu '=' xảy ra khi \(x+\frac12=0\)
=>\(x=-\frac12\)
e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)
=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)
Dấu '=' xảy ra khi \(x-\frac14=0\)
=>\(x=\frac14\)
f: \(3x^2-x+19\)
\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)
=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)
=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)
Dấu '=' xảy ra khi x-1/6=0
=>x=1/6
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)
x=\(\dfrac{-2}{5}\)
a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)
b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)
\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)
a: \(\dfrac{0.4}{x}=\dfrac{x}{0.9}\)
nên \(x^2=\dfrac{9}{25}\)
=>x=3/5 hoặc x=-3/5
b: \(\dfrac{26}{2x-1}=13\dfrac{1}{3}:1\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{26}{2x-1}=\dfrac{40}{3}:\dfrac{4}{3}=10\)
=>2x-1=13/5
=>2x=18/5
hay x=9/5
c: \(\Leftrightarrow\dfrac{2}{3}:\left(6x+7\right)=\dfrac{1}{5}:\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(6x+7\right)=\dfrac{1}{6}\)
=>6x+7=4
=>6x=-3
hay x=-1/2
d: \(\dfrac{37-x}{x+13}=37\)
=>37(x+13)=37-x
=>37x+481=37-x
=>38x=-444
hay x=-222/19
a) 3.2x+(-1.2)x+2.7= -4.9
⇔(3.2-1.2)x= -4.9-2.7
⇔2x= -7.6
⇔x= -7.6:2
⇔x= -3.8
b) \(\dfrac{37-x}{x+13}\)=\(\dfrac{3}{7}\)
⇔(37-x)*7=(x+13)*3
⇔259-7x=3x+39
⇔259-39=3x+7x
⇔220=10x
⇔x= 220:10
⇔x=22