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TH1 : \(91-3x< 7+x\Rightarrow3x+x>91-7\Rightarrow4x>84\Rightarrow x>21\left(1\right)\)
TH2 : \(7+x\ge64\Rightarrow x\ge57\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow x\ge57\)
91 - 3\(x\) < 7 + \(x\) ≥ 64
⇒ \(\left\{{}\begin{matrix}91-3x< 7+x\\7+x\ge64\end{matrix}\right.\)
\(\left\{{}\begin{matrix}7+x+3x>91\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>91-7\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>84\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>84:4\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>21\\x\ge57\end{matrix}\right.\)
\(x\ge\) 57
\(\sqrt{64}+2.\sqrt{\left(-3\right)^2}-7\sqrt{1,69}+\dfrac{3.5}{4}\)
\(=8+2.3-7.1,3+\dfrac{15}{4}\)
\(=8+6-9,1+\dfrac{15}{4}\)
\(=\dfrac{49}{10}+\dfrac{15}{4}\)
\(=\dfrac{98}{20}+\dfrac{75}{20}=\dfrac{173}{20}\)
1/
$(x-1)^{x+10}=(x-1)^{x+8}$
$\Rightarrow (x-1)^{x+10}-(x-1)^{x+8}=0$
$\Rightarrow (x-1)^{x+8}(x^2-1)=0$
$\Rightarrow (x-1)^{x+8}=0$ hoặc $x^2-1=0$
Nếu $(x-1)^{x+8}=0\Rightarrow x-1=0\Rightarrow x=1$
Nếu $x^2-1=0\Rightarrow x^2=1=1^2=(-1)^2\Rightarrow x=1$ hoặc $x=-1$
Vậy $x=1$ hoặc $x=-1$
2/
$1^3+2^3+3^3+...+10^3=(x+1)^2$
Ta có công thức quen thuộc:
$1^3+2^3+...+n^3=(1+2+...+n)^2=\frac{[n(n+1)]^2}{4}$
Bạn có thể xem cm tại đây:
https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/
Khi đó:
$1^3+2^3+...+10^3=(x+1)^2$
$\Rightarrow \frac{[10(10+1)]^2}{4}=(x+1)^2$
$\Rightarrow 3025=(x+1)^2$
$\Rightarrow x+1=55$ hoặc $x+1=-55$
$\Rightarrow x=54$ hoặc $x=-56$
Đáp ấn:D
`(3x-1)^2=64`
`<=>` $\left[ \begin{array}{l}3x-1=8\\3x-1=-8\end{array} \right.$
`<=>` $\left[ \begin{array}{l}3x=9\\3x=-7\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=3\\x=-\dfrac73\end{array} \right.$
a) \(64^x:16^x=256\)
\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)
\(\Rightarrow2^{6x}:2^{4x}=2^8\)
\(\Rightarrow2^{6x-4x}=2^8\)
\(\Rightarrow2^{2x}=2^8\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
b) \(\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)
\(\Rightarrow-7^{4-x}=-7\)
\(\Rightarrow7^{4-x}=7\)
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
c) \(\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)
\(\Rightarrow\left(-4\right)^x=-4\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)
\(\Rightarrow x=1\)
\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)
\(b,\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow7^x=-2401:\left(-7\right)\)
\(\Rightarrow7^x=343\)
\(\Rightarrow7^x=7^3\)
\(\Rightarrow x=3\)
\(c,\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)
\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)
\(\Rightarrow x=-1\)
#\(Toru\)
\(91-3.\left(7+x\right)=64\)
\(3.\left(7+x\right)=91-64\)
\(3.\left(7+x\right)=27\)
\(7+x=27:3\)
\(7+x=9\)
\(x=9-7\)
\(x=2\)
Bài này mình đã giải cho bạn rồi, xem lại nhé.