2 mũ 4 x X - 3 mũ 2 x X = 145 -255 :51
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\(2^4.x-3^2.x=145-225:51\)
\(16x-9x=145-\frac{75}{17}\)
\(7x=\frac{2390}{17}\)
\(x=\frac{2390}{119}\)
tk mình nhoa bạn
Ta có:
24.x-32.x=145-225:51
x(24-32)=145-225:51
Sai đề(225:51ko chia dc)
a)145-5[8^2-4:15]=145-5[64-4/15]=145-5.956/15=145-956/3=-521/3
b)13^2-3[15-16^2]=13^2-3.-241=169--723=892
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
2^4 .x - 3^2 .x = 145 - 255 : 51
16.x - 9.x = 145 - 5
16.x - 9.x = 140
(16 - 9).x = 140
7x = 140
x = 140 : 7
x = 20
a) 4(x+41)=7
x+41=7/4
x=7/4-41
x=157/4
b) có gì đó sai sai-.-
\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)
\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)
\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)
\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)
\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)
\(g\left(x\right)=-2x^4+4x^3+x-5\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)
`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`
`= 4x^4 + x^3 -4x^2 + x -1`
\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)
`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`
`= -2x^4 + 4x^3 +x - 5`
\(2^4.x-3^2.x=145-255:51.\)
\(2^4.x-3^2.x=145-5\)
\(2^4.x-3^2.x=140\)
\(\Leftrightarrow x.\left(2^4-3^2\right)=140\)
\(\Rightarrow x.7=140\)
\(x=140:7\)
\(x=20\)