(x-1)(x-2)(x-3)(x-4)=15
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1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
a) x* 1/5 - x* 2 * 2/15 = 4
x * 1/5 - x * 2 = 4 : 2/15
x * 15 - x * 2 = 30
x * ( 15 - 2 ) = 30
x * 13 = 30
x = 30 : 13
x = 30/13
b) ( x + x * 2 ) - 3 : 1/2 = 8
( x * 1 + x * 2 ) - 3 = 8 x 1/2
( x * 1 + x * 2 ) - 3 = 4
x * 1 + x * 2 = 4 + 3
x * 1 + x * 2 = 7
x * ( 1 + 2 ) = 7
x * 3 = 7
x = 7 : 3
x = 7/3
c)5 +( x * 1/2 - 4 ) = 15
x * 1/2 - 4 = 15 - 5
x * 1/2 - 4 = 10
x * 1/2 = 10 + 4
x * 1/2 = 14
x = 14 : 1/2
x = 7
a)x*1/5-x*2*2/15=4
x*1/5-x*4/15=4
x*1/5-4/15=4
x*(-1/15)=4
x=4:-1/15
x=-60
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
\(a,\left(x+1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)\(\Leftrightarrow x^3+3x^2+3x+1+8-x^3+3x^2+6x-17=0\)\(\Leftrightarrow6x^2+9x-8=0\)
\(\Leftrightarrow x^2+\dfrac{3}{2}x-\dfrac{4}{3}=0\)
\(\Leftrightarrow\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{16}-\dfrac{4}{3}=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2=\dfrac{91}{48}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\sqrt{\dfrac{91}{48}}\\x+\dfrac{3}{4}=-\sqrt{\dfrac{91}{48}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\\x=-\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{273}}{12}\\x=-\dfrac{9+\sqrt{273}}{12}\end{matrix}\right.\)
b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x-15=0\)
\(\Leftrightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
Bài này là của lớp 6 mà, hay là bạn học lớp 1 ko
(x-1)(x-2)(x-3)(x-4)=15
<=> (x-1)(x-4)(x-2)(x-3)=15
<=> (x2-4x-x+4)(x2-2x-3x+6)=15
<=> (x2-5x+4)(x2-5x+6)=15 (*)
Đặt t=x2-5x+5
(*) <=> (t+1)(t-1)=15
<=> t2-1=15
<=> t2=16
<=> t=4 hoặc t=-4
<=> x2-5x+5=4 hoặc x2-5x+5=-4
+) x2-5x+5=4
<=> x2-5x+1=0
\(\Leftrightarrow x^2-2.\frac{5}{2}x+\frac{25}{4}=\frac{21}{4}\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)
.................(tự làm)
\(x=\orbr{\begin{cases}\frac{5+\sqrt{21}}{2}\\\frac{5-\sqrt{21}}{2}\end{cases}}\)
+) x2-5x+5=-4
<=> x2-5x+9=0
\(\Leftrightarrow x^2-2.\frac{5}{2}x+\frac{25}{4}=-\frac{11}{4}\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=-\frac{11}{4}\)
vì (x-5/2)2>0 => pt vô nghiệm
vậy \(x=\orbr{\begin{cases}\frac{5+\sqrt{21}}{2}\\\frac{5-\sqrt{21}}{2}\end{cases}}\)