23-3x X = 13
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Ta có:
\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{7}{13\times20}+\frac{3}{20\times23}\)
=>\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\)
=>\(\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
25-{14-[(2-x)-(x+13)+23]}=62-{8-[(2x-10)-(29-3x)+8]}
<=>25-14+2-x-x-13+23=62-8+2x-10-29+3x+8
<=>13-x-x-13+23=2x+3x+8-10-29
<=>-x-x+23=2x+3x+(-31)
<=>23-(x+x)=2x+2x-31
<=>23-2x=2x+3x-31
<=>23+31=2x+2x+3x
<=>54=x(2+2+3)
<=>54=x.7
=>x=54/7
Vậy x=54/7
<=>54=x.7
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T a c ó : A = B ⇔ ( x + 1 ) 3 – ( x – 2 ) 3 = ( 3 x – 1 ) ( 3 x + 1 ) ⇔ x 3 + 3 x 2 + 3 x + 1 – x 3 + 6 x 2 – 12 x + 8 = 9 x 2 – 1 ⇔ x 3 – x 3 + 3 x 2 + 6 x 2 – 9 x 2 + 3 x – 12 x = - 1 – 1 – 8 ⇔ - 9 x = - 10 ⇔ x = 10 / 9 V ậ y v ớ i x = 10 / 9 t h ì A = B .
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\(C=\left(23-x\right)\left(3x+5\right)+13\)
\(=69x+115-3x^2-5x+13\)
\(=-3x^2+64x+128\)
\(=-3\left(x^2-\dfrac{64}{3}x+\dfrac{1024}{9}\right)+\dfrac{1408}{3}\)
\(=-3\left(x-\dfrac{32}{3}\right)^2+\dfrac{1408}{3}\le\dfrac{1408}{3}\)
Vậy \(Max_C=\dfrac{1408}{3}\)
Để \(C=\dfrac{1408}{3}\) thì \(x-\dfrac{32}{3}=0\Rightarrow x=\dfrac{32}{3}\)
d, \(D=\left(2-3x\right)\left(3x+5\right)-7\)
\(=6x+10-9x^2-15x-7\)
\(=-9x^2-9x+3\)
\(=-9\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{21}{4}\)
\(=-9\left(x-\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy \(Max_D=\dfrac{21}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
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a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.
b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.
c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.
d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(60+x-34=43\Rightarrow x=43+34-60=17\)
\(23x-6.23+69x=230\)
\(\Rightarrow23x+69x=230+6.23\)
\(\Rightarrow92x=368\Rightarrow x=368:92\Rightarrow x=4\)
\(\left(15.24-x\right):0,25=100:0,25\)
\(\Rightarrow\left(360-x\right)=100:0,25.0,25\)
\(\Rightarrow360-x=100\Rightarrow x=360-100=260\)
\(\left(7x+8\right):5=10\Rightarrow\left(7x+8\right)=10.5\)
\(\Rightarrow7x+8=50\Rightarrow7x=50-8\Rightarrow7x=42\Rightarrow x=42:7=6\)
\(280:\left(7+3x\right)=4\Rightarrow\left(7+3x\right)=280:4\)
\(\Rightarrow7+3x=70\Rightarrow3x=70-7\Rightarrow3x=63\Rightarrow x=63:3=21\)
\(\left(x+5\right)x19:13=57\)
\(\Rightarrow\left(x+5\right)x19=57.13\)
\(\Rightarrow\left(x+5\right)=741:19\Rightarrow x+5=39\Rightarrow x=39-5=34\)
\(4\left(36-4x\right)=64\)
\(\Rightarrow\left(36-4x\right)=64:4\)
\(\Rightarrow36-4x=16\)
\(\Rightarrow4x=36-16\Rightarrow4x=20\Rightarrow x=20:4=5\)
\(\left(7,6:1,9\right)x=3,2\)
\(\Rightarrow4x=3,2\Rightarrow x=3,2:4=0,8\)
\(\left(x:2+50\right):5=12\)
\(\Rightarrow x:2+50=12.5\)
\(\Rightarrow x:2=60-50\Rightarrow x:2=10\Rightarrow x=10.2=20\)
có 2 x à bé
cái nào là x cái nào là nhân thế??????