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30 tháng 7 2020

a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)

c) \(73^2-13^2-10^2+20.13\)

\(=73^2-\left(13^2+10^2-20.13\right)\)

\(=73^2-\left(13^2-2.13.10+10^2\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2\)

\(=\left(73-3\right)\left(73+3\right)\)

\(=70.76\)

\(=5320\)

d)Viết đề = công thức trực quan hộ mình

a) Ta có: \(85^2-15^2\)

\(=\left(85-15\right)\left(85+15\right)\)

\(=70\cdot100=7000\)

b) Ta có: \(93^3+21\cdot93^2+3\cdot49\cdot93+343\)

\(=93^3+3\cdot93^2\cdot7+3\cdot93+7^2+7^3\)

\(=\left(93+7\right)^3\)

\(=100^3=1000000\)

c) Ta có: \(73^2-13^2-10^2+20\cdot13\)

\(=73^2-\left(13^2+10^2-20\cdot13\right)\)

\(=73^2-\left(13^2-2\cdot13\cdot10+10^2\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2=\left(73-3\right)\left(73+3\right)\)

\(=70\cdot76=5320\)

16 tháng 8 2020

a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)

b) \(93^3+21.93^2+3.49.93+343\)

\(=93^3+3.7.93^2+3.7^2.93+7^3\)

\(=\left(93+7\right)^3\)

\(=100^3=1000000\)

c) \(73^2-13^2-10^2+20.13\)

\(=73^2-\left(13^2+10^2-20.13\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2\)

\(=\left(73+3\right)\left(73-3\right)\)

\(=76.70=5320\)

d) Viết = Latex hộ mình

20 tháng 10 2020

Nốt câu c

\(C=\frac{97^3+83^3}{180}-97.83\)

\(=\frac{\left(97+83\right)\left(97^2-97.83+83^3\right)}{180}-97.83\)

\(=\frac{180.\left(97^2-97.83+83^2\right)}{180}-97.83\)

\(=97^2-97.83+83^2-97.83\)

\(=\left(97-83\right)^2=14^2=196\)

20 tháng 10 2020

1.

a) \(2\left(x+3\right)-x^2-3x=0\)

\(2\left(x+3\right)-x\left(x+3\right)=0\)

\(\left(x+3\right)\left(2-x\right)=0\)

\(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

b) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

\(3x\left(x+2\right)=0\)

\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c) \(x^2-4x+3=0\)

\(x\left(x-1\right)-3\left(x-1\right)=0\)

\(\left(x-1\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

2, a) \(B=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)-2x\)

\(B=8x^3+27-8x^3+2-2x=29-2x\)

Tại x = 3

Thì B = 29 - 6 = 23

10 tháng 2 2018

10)   \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)

\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)   (vì  1/86 + 1/85 + 1/84 + 1/83 + 1/4  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy....

4 tháng 2 2020

a) Đặt x -3 = a

<=> a(a+2)(a+8)(a+10) - 297=0

<=> \(\left[a\left(a+10\right)\right]\left[\left(a+2\right)\left(a+8\right)\right]\)-297=0

<=> \(\left(a^2+10a\right)\left(a^2+10a+16\right)-297=0\)

Đặt \(a^2+10a=b\)

\(b^2+16b-297=0\)

\(\Rightarrow\left[{}\begin{matrix}b=11\\b=-27\end{matrix}\right.\)\(b=11\Rightarrow\left[{}\begin{matrix}a=1\\a=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

b= -27 \(\Rightarrow a=\varnothing\Rightarrow x=\varnothing\)

4 tháng 2 2020

b) bấm máy ra nhân tử chung :D

c)

\(\Leftrightarrow\left(\frac{1927-X}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+...=0\)

\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

<=> x = 2018

d) \(\Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-3\right)=0\)

giống câu c

9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)

=>x-99=0

hay x=99

7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)

=>x+100=0

hay x=-100

8:

Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\) 

\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)

=>200-x=0

hay x=200

5 tháng 2 2018

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)            (vì 1/75 + 1/70 - 1/65 - 1/60  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy.....

7 tháng 2 2018

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)    (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy...

AH
Akai Haruma
Giáo viên
5 tháng 1 2019

Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)

\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)

\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)

AH
Akai Haruma
Giáo viên
5 tháng 1 2019

Bài 2:

\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)

\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)

15 tháng 3 2022

mình thấy đề hơi sai sai sao á bạn đoạn \(\dfrac{x-4}{93}\) á bạn