ủa cái này hình như bn lạc lõn gì đấy trả lời r mà ?

\(\Leftrightarrow80\left(2x-9\right)=240\times39\)

\(\Leftrightarrow160x-720=9360\)

\(\Leftrightarrow160x=9360+720\)

\(\Leftrightarrow160x=10080\)

\(\Leftrightarrow x=63\)

a. \(M=-x^4y^4\)

b.\(-\left(2^2\right).\left(-1\right)^2\)=(-2)

a)\(M=\left(\dfrac{2}{3}\cdot\left(-\dfrac{3}{2}\right)^2\right).\left(x.x^2\right)\left(y^2.y^2\right)=\dfrac{3}{2}.x^3y^4\)

hệ số : 3/2 

biến :\(x^3y^4\)

b) thay x=2 ; y=-1 và M ta đc

\(M=\dfrac{3}{2}.2^3.\left(-1\right)^4=\dfrac{3}{2}\cdot8.1=\dfrac{24}{2}=12\)

Ta có :

\(\dfrac{x}{10}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{3}\Leftrightarrow\dfrac{y}{10}=\dfrac{z}{15}\)

\(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)

\(\Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{33}{7}\Leftrightarrow x=\dfrac{660}{7}\\\dfrac{y}{10}=\dfrac{33}{7}\Leftrightarrow y=\dfrac{330}{7}\\\dfrac{z}{15}=\dfrac{33}{7}\Leftrightarrow z=\dfrac{495}{7}\end{matrix}\right.\)

Vậy .....

Cảm ơn bạn nhek

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

thank bn nha

a: =>4x^2-4x+1+7>4x^2+3x+1

=>-4x+8>3x+1

=>-7x>-7

=>x<1

b: \(\Leftrightarrow12x+1>=36x+12-24x-3\)

=>1>=9(loại)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

a: ĐKXĐ: x∉{1;-1}

Ta có: \(Q=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right)\cdot\left(\frac{1-x^2}{2}\right)^2\)

\(=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x+2}{\left(x+1\right)^2}\right)\cdot\frac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

\(=\frac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\cdot\frac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

\(=\frac{x^2-x-2-\left(x^2+x-2\right)}{1}\cdot\frac{x-1}{4}=\frac{-2x}{4}\cdot\left(x-1\right)=-\frac12x\cdot\left(x-1\right)\)

b: Q=4-5x

=>Q=5x+4

=>\(-\frac12\left(x^2-x\right)=5x+4\)

=>\(-\left(x^2-x\right)=10x+8\)

=>\(x^2-x=-10x-8\)

=>\(x^2-x+10x+8=0\)

=>\(x^2+9x+8=0\)

=>(x+1)(x+8)=0

=>\(\left[\begin{array}{l}x+1=0\\ x+8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\left(loại\right)\\ x=-8\left(nhận\right)\end{array}\right.\)

Bài 1 :

Đặt :

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=3k\\3y=4k\\4z=5k\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3k}{2}\\y=\dfrac{4k}{3}\\z=\dfrac{5k}{4}\end{matrix}\right.\)

Thay vào \(x+y+z=49\) ta được :

\(\dfrac{3k}{2}=\dfrac{4k}{3}=\dfrac{5k}{4}=49\)

\(\Leftrightarrow\dfrac{18k+16k+15k}{12}=\dfrac{588}{12}\)

\(\Leftrightarrow49k=588\)

\(\Leftrightarrow k=12\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3.12}{2}=18\\y=\dfrac{4.12}{3}=16\\z=\dfrac{5.12}{4}=15\end{matrix}\right.\)

Vậy ....

Bài1:

Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}\)

Áp dụng t/c của dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}=\dfrac{x+y+z}{90+80+75}=\dfrac{49}{245}=\dfrac{1}{5}\)

=>x=18;b=16;c=15

Vậy...

a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\) 

b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\) 

     \(\dfrac{x}{3}=\dfrac{-14}{15}\) 

\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)

\(x=\dfrac{-2}{7}+\dfrac{9}{7}\) 

\(x=1\)