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Ta có: \(x^2-y^2-2x-2y=\left(x^2-y^2\right)-\left(2x+2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

Số đó là : 76

Tích đúng là : 2660

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Ta có: \(x+2x+3x+....+20x=210\)

\(\Leftrightarrow x\left(1+2+3+...+20\right)=210\)

\(\Leftrightarrow\dfrac{20\cdot\left(20+1\right)}{2}x=210\)

\(\Rightarrow x=\dfrac{210\cdot2}{20\cdot\left(20+1\right)}=1\)

Vậy \(x=1\)

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a) Gọi x,(x+1),(x+2) làn lượt là ba số tự nhiên liên tiếp

Theo đề bài ta có: \(x\left(x+1\right)\left(x+2\right)=1716\)

\(\Leftrightarrow x^3+3x^2+2x=1716\)

\(\Rightarrow x^3+3x^2+2x-1716=0\)

\(\Leftrightarrow x^3-11x^2+14x^2-154x+156x-1716=0\)

\(\Leftrightarrow x^2\left(x-11\right)+14x\left(x-11\right)+156\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-11\right)\left(x^2+14x+156\right)=0\)

Vì \(x^2+14x+156=\left(x+7\right)^2+107>0\)

Nên \(x-11=0\Rightarrow x=11\)

Vậy ba số liên tiếp đó là: 11,12,13

b) Gọi b số Chẵn liên tiếp đó là: 2x-2,2x,2x+2

Theo đề bài có: \(2x\left(2x-2\right)\left(2x+2\right)=5760\)

\(\Leftrightarrow8x^3-8x=5760\)

\(\Rightarrow8x^3-8x-5760=0\)

\(\Leftrightarrow8x^3-72x^2+72x^2-648x+640x-5760=0\)

\(\Leftrightarrow8x^2\left(x-9\right)+72x\left(x-9\right)+640\left(x-9\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(8x^2+72x+640\right)=0\)

Vì \(8x^2+72x+640>0\) nên \(x-9=0\Rightarrow x=9\)

Vậy ba số chẵn liên tiếp đó là 16,18,20

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Ta có: \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=3\)

\(\Leftrightarrow\dfrac{xyz}{z^2}+\dfrac{xyz}{y^2}+\dfrac{xyz}{x^2}=3\)

\(\Leftrightarrow xyz\left(\dfrac{1}{z^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2}\right)=3=1\cdot3=3\cdot1\) (Vì x,y,z dương)

TH1: \(\left\{{}\begin{matrix}xyz=1\\\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\end{matrix}\right.\) \(\Rightarrow x=y=z=1\) (TM)

TH2: \(\left\{{}\begin{matrix}xyz=3\\\dfrac{1}{x^2}+\dfrac{1}{y^1}+\dfrac{1}{z^2}=1\end{matrix}\right.\) \(\Rightarrow x=y=z=\varnothing\)

Vậy x=y=z=1

/-x-2/=5

/-x/=5+2

/-x/=7

vậy x=7 hoặc x=-7

tick cho mk nha bạn

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Đặt \(A=1+4+7+...+97+100\)

Ta có số các số hạng trong A là: \(\dfrac{\left(100-1\right)}{3}+1=34\) (số hạng)

Do đó \(A=\dfrac{\left(100+1\right)\cdot34}{2}=1717\)

Vậy \(1+4+7+...+100=1717\)

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1) Theo đề bài ta có: \(\left\{{}\begin{matrix}a=5b+7\\a-b=215\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5b+7\\5b+7-b=215\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=267\\b=52\end{matrix}\right.\)

Vậy a=267 và b=52

2) Theo đề bài có: \(\left\{{}\begin{matrix}a=4b+11\\a+b=215\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=4b+11\\4b+11+b=215\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=174.2\\b=40.8\end{matrix}\right.\)

Vậy a=174.2 và b=40.8

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Ta có: \(A=\sqrt{x}+\sqrt{y}\)

\(\Rightarrow A^2=x+y+2\sqrt{xy}=1+2\sqrt{xy}\) (vì x+y=1)

\(\Leftrightarrow A^2\ge1\) (Vì \(2\sqrt{xy}\ge0\))

\(\Rightarrow A\ge\sqrt{1}=1\)

Vậy MinA=1 \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)