tính nhanh

35 x 11 x 0,1 + 0,25 x 100 x 3 : 0,4 - 7,5  =  218,5

1) Ta có: \(3x\left(2-5x\right)+35x-14=0\)

\(\Leftrightarrow3x\left(2-5x\right)+7\left(5x-2\right)=0\)

\(\Leftrightarrow-3x\left(5x-2\right)+7\left(5x-2\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(-3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=2\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{5};\dfrac{7}{3}\right\}\)

2) Ta có: \(4x-6+5x\left(3-2x\right)=0\)

\(\Leftrightarrow2\left(2x-3\right)-5x\left(2x-3\right)=0\)

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

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

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

/ là gì vậy em..giải thích đi chị trả lời cho

39x49 - 1 trên 35x48 + 34 

Tính bằng cách thuật tiện nhất. Ạ

= 955/875 em

chắc chắn đúng k chị nha

ta có : \(35x-14y+2^9-1=35x-14x+511\)

\(=7\left(5x-2y+73\right)⋮7\forall x\overset{.}{,}y\left(đpcm\right)\)

\(6x-1=2x+3\\ \Rightarrow4x=4\\ \Rightarrow x=1\)

\(5x\left(x-2\right)+\left(2x^4+10x^3-4x^2\right):x^2\\ =5x^2-10x+2x^2\left(2x^2+5x-4\right):x^2\\ =5x^2-10x-2\left(2x^2+5x-4\right)\\ =5x^2-10x-4x^2-10x+8\\ =x^2-20x+4\)

\(\left(x+2\right)^2-2x-4=0\\ \Rightarrow\left(x+2\right)^2-2\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(x+2-2\right)=0\\ \Rightarrow x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Ta có : \(6x^4-35x^3+62x^2-35x+6=0\)

=> \(6x^4-3x^3-32x^3+16x^2+46x^2-23x-12x+6=0\)

=> \(3x^3\left(2x-1\right)-16x^2\left(2x-1\right)+23x\left(2x-1\right)-6\left(2x-1\right)=0\)

=> \(\left(3x^3-16x^2+23x-6\right)\left(2x-1\right)=0\)

=> \(\left(3x^3-x^2-15x^2+5x+18x-6\right)\left(2x-1\right)=0\)

=> \(\left(x^2\left(3x-1\right)-5x\left(3x-1\right)+6\left(3x-1\right)\right)\left(2x-1\right)=0\)

=> \(\left(x^2-5x+6\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x^2-2x-3x+6\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x\left(x-2\right)-3\left(x-2\right)\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x-3\right)\left(x-2\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left[{}\begin{matrix}x-3=0\\x-2=0\\3x-1=0\\2x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=2\\x=\frac{1}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{2,3,\frac{1}{2},\frac{1}{3}\right\}\)

Nhận thấy \(x=0\) ko là nghiệm, chia 2 vế của pt cho \(x^2\)

\(6x^2+\frac{6}{x^2}-35x-\frac{35}{x}+62=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(6\left(t^2-2\right)-35t+62=0\)

\(\Leftrightarrow6t^2-35t+50=0\Rightarrow\left[{}\begin{matrix}t=\frac{5}{2}\\t=\frac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\x+\frac{1}{x}=\frac{10}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-5x+2=0\\3x^2-10x+3=0\end{matrix}\right.\)