\(\left(x-1\right)^5=-32\)

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

\(\Rightarrow x-1=-2\)

\(\Rightarrow x=-2+1\)

\(\Rightarrow x=-1\)

(x-1)⁵= -32

=>(x-1)⁵=(-2)⁵

=> x-1 = -2

=> x = -2 +1

=> x = -1.

(x-1)^5= -32

(x-1)^5= -1^5

x-1 = -1

x= -1+1

x = 0

Gần chục năm rồi đấy

\(\frac{9}{7}x\frac{14}{3}+5\)

=\(\frac{9}{7}x\frac{14}{3}=6+5\)

=\(6+5=11\)

k mình nha

lớp 1 ý

\(\frac{9}{7}\) \(\times\) \(\frac{14}{3}\) +     5

= \(\frac{9\times14}{7\times3}\) +      5

= 6     +      5 

= 11

a, -75 . (-27).(-4)

= 2025 .(-4)

=-8100

b. 1x2x3x4x5x(-10)

=120 x (-10)

=-1200

1.

\(\Delta=m^2-4\left(2m-5\right)=\left(m-4\right)^2+4>0;\forall m\)

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=2m-5\end{matrix}\right.\)

Để biểu thức đề bài xác định \(\Rightarrow m\ne-2\)

\(A=\frac{x_1x_2}{x_1+x_2+2}=\frac{2m-5}{m+2}=2-\frac{9}{m+2}\)

\(A\in Z\Rightarrow\frac{9}{m+2}\in Z\Rightarrow m+2=Ư\left(9\right)\)

\(\Rightarrow m+2=\left\{-9;-3;-1;1;3;9\right\}\)

\(\Rightarrow m=\left\{-11;-5;-3;-1;1;7\right\}\)

2.

Hệ pt tọa độ giao điểm A của d1 và d2: \(\left\{{}\begin{matrix}x+y=2\\-2x+y=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) \(\Rightarrow A\left(1;1\right)\)

Để 3 đường thẳng đồng quy \(\Leftrightarrow\) d qua A

\(\Leftrightarrow1=\left(m-2\right).1+m+3\Rightarrow2m=0\Rightarrow m=0\)

b/ Gọi \(B\left(x;y\right)\) là điểm cố định mà d luôn đi qua

\(\Leftrightarrow y=\left(m-2\right)x+m+3\) ; \(\forall m\)

\(\Leftrightarrow m\left(x+1\right)+\left(-2x-y+3\right)=0\)

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

Vậy d luôn đi qua \(B\left(-1;5\right)\)

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

\(\Rightarrow\left(5x^2+3x-2\right)^2-\left(4x^2-3x-2\right)^2=0\)

\(\Rightarrow\left[\left(5x^2+3x-2\right)-\left(4x^2-3x-2\right)\right]\left[\left(5x^2+3x-2\right)+\left(4x^2-3x-2\right)\right]=0\)

\(\Rightarrow\left(5x^2+3x-2-4x^2+3x+2\right)\left(5x^2+3x-2+4x^2-3x-2\right)=0\)

\(\Rightarrow\left(x^2+6x\right)\left(9x^2-4\right)=0\)

\(\Rightarrow x\left(x+6\right)\left[\left(3x\right)^2-2^2\right]=0\)

\(\Rightarrow x\left(x+6\right)\left(3x-2\right)\left(3x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\\3x-2=0\\3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\3x=2\\3x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

x/8=1/y<=>xy=1.8=8(nhân chéo)

=>x,y E Ư(8)

do đó

+)x=1=>y=8

+)x=8=>y=1

+)x=2=>y=4

+)x=4=>y=2

+)x=-1=>y=-8

+)x=-8=>y=-1

+)x=-2=>y=-4

+)x=-4=>y=-2

vậy ....

đề có bấy nhiu thui hả

tìm x,y biết\(\frac{x}{8}=\frac{1}{y}\)