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8 tháng 7 2018

\(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}=\frac{\left(-5\right)^5.16}{5^4.16}=\frac{\left(-5\right)^5}{5^4}=-5\)

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{14^{1005}.5^{1005}.5}{2^{1004}.2^3.35^{1004}}=\frac{\left(14.5\right)^{1005}.5}{\left(2.35\right)^{1004}.2^3}=\frac{70^{1005}.5}{70^{1004}.2^3}=\frac{70.5}{8}=\frac{350}{8}=\frac{175}{4}\)

Chúc bạn học tốt!

8 tháng 7 2018

Toán lớp 6 ? 

\(\frac{\left(-5\right)^2\left(-5\right)^3.16}{5^4.\left(-2\right)^4}=\frac{\left(-5\right)^5.2^4}{5^4.2^4}=-5\)

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{14^{1005}.5^{1005}.5}{2^{1004}.35^{1004}.2^3}=\frac{\left(14.5\right)^{1005}.5}{\left(2.35\right)^{1004}.8}=\frac{70^{1005}.5}{70^{1004}.8}=\frac{70.5}{8}=\frac{175}{4}\)

10 tháng 12 2021

a: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

10 tháng 12 2021

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

10 tháng 12 2021

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

10 tháng 12 2021

Sao lm mỗi câu a, c thế

Câu c viết dấu hoặc đi

1 tháng 7 2015

tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho

22 tháng 7 2017

Đề là gì v bn?

30 tháng 11 2016

\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

\(x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(x-2\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

\(x\left(x-1\right)+5x-5=0\)

\(x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(x+5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)

\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

30 tháng 11 2016

Cảm ơn

 

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

23 tháng 4 2023
cho A=1/101+1/102+1/103+...+1/199+1/200 chứng minh 1/2 <A<1

a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)

7 tháng 12 2016

 

Tính
a) ( 2x + 5 )2 + ( 2x + 5 )2 - 2(2x + 3 ) (2x + 5 )

=> Sai đề

7 tháng 12 2016

 

b) ( x - 3 ) ( x + 5 ) - ( x - 3 ) 2

=(x-3)[(x+5)-(x-3)]

=(x-3)(x+5-x+3)

=(x-3).8

13 tháng 11 2021

(x + 2)2 - 2x - 4 = 0

<=> (x + 2)2 - 2(x + 2) = 0

<=> (x + 2)(x + 2 - 2) = 0

<=> x(x + 2) = 0

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

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