Question

A math question most computers can't solve but you can.
${( {x}^{2} - 7x + 11) }^{ {x}^{2} - 13x + 42 } = 1$
please Don't scam .​

Answer 1

Answer:

Firstly,

we know that,

anything to the power zero (0) = 1 , that is

$\sf {a}^{0} = 1 \: \: \: where \: \: a \neq 0$

That's why, we can also write that,

$\sf \: 1 = {( {x}^{2} - 7x + 11 })^{0}$

and

$\sf \: \:if \: \: {a}^{m} = {a}^{n} \: \: \: \: then \: \: m = n$

Now,

Given,

$\sf \: {( {x}^{2} - 7x + 11 })^{ { {x}^{2} - 13x + 42}^{} } = 1 \\ \sf \implies \: {( {x}^{2} - 7x + 11 })^{ { {x}^{2} - 13x + 42}^{} } = {( {x}^{2} - 7x + 11)}^{0} \\ \sf \implies \: {x}^{2} - 13x + 42 = 0 \\ \sf \implies \: {x}^{2} - 7x - 6x + 42 = 0 \\ \sf \implies \: x(x - 7) - 6(x - 7) = 0 \\ \sf \implies \: (x - 7)(x - 6) = 0 \\ \therefore \sf \: x = 7 \: \: \: or \: \: \: 6$

Answer 2

$\color{darkred}question = {( {x}^{2} - 7x + 11) }^{ {x}^{2} - 13x + 42 } = 1 \\ \\ \color{darkkhaki}solution \\ \color{darkgreen}formula = {a}^{0} = 1. \\ \\ \color{green}{( {x}^{2} - 7x + 11) }^{ {x}^{2} - 13x + 42 } = 1 \\ \ \\ \small\color{green} {( {x}^{2} - 7x + 11) }^{ {x}^{2} - 13x + 42 } =( {x}^{2} - 7x + 11) {}^{0} \\ \\ \color{green} {x}^{2} - 13x + 42 = 0 \\ \\ \color{green} {x}^{2} - (7 + 6)x + 42 = 0 \\ \\ \color{green} {x}^{2} - 7x - 6x + 42 = 0 \\ \\ \color{green}x(x - 7) - 6(x - 7) = 0 \\ \\ \color{green}(x - 7)(x - 6) = 0 \\ \\ \color{red} || \: \: \color{blue}x - 7 = 0 \\ \color{blue}x = 7\color{red} \: \: || \\ \\\color{red} || \color{green}x - 6 = 0 \\ \color{green}x = 6 \: \: \color{red} || \\ \\ \color{greenyellow}zerose \: of \: the \: polynomial \\\color{darkgreen} x = 7 \: , \: x = 6 \\ \\\color{orange} hope \: its \: help \: you ❤\\ \\ \small \color{red} its \: ᭄亗 乄 ＭꫝղᎥនh 乄 亗✯❤࿐$