TÌM X BIẾT 10 + (2X-1)MŨ 2 :3 =13
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Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
125 : 5^2 . (x-3) = 3
125 : 25 .(x-3) = 3
5 (x-3) =3
x-3 = 3 : 5
Đề bài sai hay sao ý?
2x -5 = 3
2x = 3+5 =8
x =4
(19 .x+2.5^2 ): 14 = (13-8)^2 - 4^2
19x + 50) : 14 = 25 - 16
(19x+50) : 14 = 9
19x+50 = 3*14 = 126
19x = 126-50 = 76
x = 76 / 19 = 4
\(a\)) \(125:5^2.\left(x-3\right)=3\)
\(5^3:5^2.\left(x-3\right)=3\)
\(5.\left(x-3\right)=3\)
\(5x-15=3\)
\(5x=3+15\)
\(5x=18\)
\(x=\frac{18}{5}\)\(.\)Vậy \(x=\frac{18}{5}\)
\(b\)) \(\left(2x-5\right)^3=27\)
\(\left(2x-5\right)^3=3^3\)
\(\Rightarrow2x-5=3\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)\(.\)Vậy \(x=4\)
\(c\))\(\left(19.x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\left(19.x+2.25\right):14=5^2-16\)
\(\left(19.x+2.25\right):14=25-16\)
\(\left(19.x+50\right):14=9\)
\(19.x+50=126\)
\(19.x=126-50\)
\(19.x=76\)
\(x=76:19\)
\(x=4\)\(.\)Vậy \(x=4\)
a) Ta có: \(\frac{x+2}{2}-\frac{2x-3}{5}=\frac{10x+13}{10}\)
\(\Leftrightarrow\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}-\frac{10x+13}{10}=0\)
Suy ra: \(5x+10-4x+6-10x-13=0\)
\(\Leftrightarrow-9x+3=0\)
\(\Leftrightarrow-9x=-3\)
hay \(x=\frac{1}{3}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-1}{x-2}-\frac{5}{x+2}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2}{\left(x+2\right)\left(x-2\right)}=0\)
Suy ra: \(x^2+x-2-5x+10-x^2=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)
Con " Nguyễn Huyền Trang " đéo biết thì trả lời làm cái l*n gì