mk nhịu

a)    \(\frac{28\times7-45\times7+7\times18}{45\times14}\)

\(=\frac{7\left(28-45+7\right)}{45\times14}\)

\(=\frac{7\times\left(-10\right)}{45\times14}=\frac{-1}{9}\)

b)   \(\frac{12.3-2.6}{4.5.6}\)

\(=\frac{2.6.3-2.6}{4.5.6}\)

\(=\frac{2.6\left(3-1\right)}{2.2.5.6}\)

\(=\frac{2.6.2}{2.2.5.6}\)\(=\frac{1}{5}\)

a)\(\frac{28\times7-45\times7+7\times18}{45\times14}\)

\(=\frac{7\times\left(28-45+18\right)}{45\times14}\)

\(=\frac{7\times1}{45\times14}\)

\(=\frac{1}{90}\)

b)\(\frac{12\times3-2\times6}{4\times5\times6}\)

\(=\frac{12\times\left(3-1\right)}{4\times5\times6}\)

\(=\frac{6\times2\times2}{2\times2\times5\times6}\)

\(=\frac{1}{5}\)

P/s em tham khảo nhé

Lời giải:

Gọi biểu thức cần rút gọn là $P$

Xét tử số: $\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3.1}+1}-\sqrt{3}$

$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{3}=|\sqrt{3}+1|-\sqrt{3}=1$

Xét mẫu số:

Ta dự đoán sẽ rút gọn được $\sqrt[3]{17\sqrt{5}-38}$

Đặt $17\sqrt{5}-38=(a+\sqrt{5})^3$ với $a$ nguyên.
$\Leftrightarrow 17\sqrt{5}-38=a^3+15a+\sqrt{5}(3a^2+5)$

$\Rightarrow 17=3a^2+5$ và $-38=a^3+15a$

$\Rightarrow a=-2$

Vậy $17\sqrt{5}-38=(-2+\sqrt{5})^3$

$\Rightarrow (\sqrt{5}+2)\sqrt[3]{17\sqrt{5}-38}=(\sqrt{5}+2)(-2+\sqrt{5})=1$

Vậy $P=\frac{1}{1}=1$

a)\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^8-1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^{16}-1\right)\)

\(=\dfrac{1}{2}3^{16}-\dfrac{1}{2}\)

b) \(48\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^4-1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^8+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^{32}-1\right)\)

\(=2.5^{32}-2\)

Tham khảo nhé~

3 chấm ở giữa để kia làm j z

Ai đó giúp tui nhaaa, đa tạ !!

Xét \(x<4\Rightarrow |x-4|=4-x\)

                    \(|x-5|=5-x\)

Biểu thức \(A=4-x+5-x=9-2x\)

Xét \(4\leq x<5 \Rightarrow |x-4|=x-4\) và \(|x-5|=5-x\) thay vào \(A=1\)

Xét \(x\geq5\Rightarrow|x-4|=x-4\) và \(|x-5|=x-5\) thay vào \(A=2x-9\)

\(|x-5|\)luôn \(\ge0\)

\(\Rightarrow\hept{\begin{cases}|x-5|=x-5\\|x-5|=-\left(x-5\right)=-x+5\end{cases}}\)

\(|x-4|\)luôn \(\ge0\)

\(\Rightarrow\hept{\begin{cases}|x-4|=x-4\\|x-4|=-\left(x-4\right)=-x+4\end{cases}}\)

Ta có các trường hợp:

\(\hept{\begin{cases}\text{|x-5|+|x-4|}=\left(x-5\right)+\left(x-4\right)=x-5+x-4=2x-9\\\text{|x-5|+|x-4|}=\left(-x+5\right)+\left(x-4\right)=-x+5+x-4=1\end{cases}}\)

\(\hept{\begin{cases}\text{|x-5|+|x-4|}=\left(-x+4\right)+\left(x-5\right)=-x+4+x-5=-1\\\text{|x-5|+|x-4|}=\left(-x+4\right)+\left(-x+5\right)=-x+4-x-5=-2x-1\end{cases}}\)

\(\left(2x-5\right)\left(2x+5\right)-\left(2x+1\right)^2=4x^2-25-4x^2-4x-1=-4x-25=\left(-4\right).\left(-2005\right)-26=8020-26=7994\)

câu đó ở đây

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\(\left[\frac{\sqrt{5}\left(1+\sqrt{5}\right)}{\sqrt{5}}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}\right].\frac{\sqrt{2}+\sqrt{5}}{1}=1+\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)=1+3=4\)

\(a,=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\\ b,=\dfrac{\left(x+y\right)^2-16}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)

a, \(\left(a+5\right)^2+2\left(a+5\right)\left(\dfrac{1}{2}-a\right)+\left(\dfrac{1}{2}-a\right)^2=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\)

b,\(\dfrac{x^2-16+2xy+y^2}{3x^2-12x+3xy}=\dfrac{\left(x^2+2xy+y^2\right)-4^2}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)