a)(x+3)2-x2+15=2x+6-2x+15=1

                        =21=1

Bạn chép sai đầu bài à

x2 là x^2 à

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

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

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

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

d) Ta có: \(x^2-6x-16=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

Bài 1 :

a, \(A=x\left(x-6\right)+10\)

=x^2 - 6x + 10

=x^2 - 2.3x+9+1

=(x-3)^2 +1 >0 Với mọi x dương

Cảm ơn bạn Vũ Anh Quân ;) ;) ;) 

bạn vừa đăng câu này r mà

a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)

\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)

\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

hay \(x\in\left\{1;-1\right\}\)

b: Ta có: \(x^2-7x+10=0\)

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

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

bạn học sd máy tính tìm nghiệm chưa?

a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)

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

d) \(x^4+7x^2-18=0\\ \Rightarrow\left(x^4-4\right)+\left(7x^2-14\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2\right)+7\left(x^2-2\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x^2=-9\left(loại\right)\end{matrix}\right.\)

\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)

a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)

\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)

\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)

b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)

\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)

\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)

\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)

c, \(4x^2-2x-1=0\)

\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)

\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)

\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)

d,\(x^4-4x^2-32=0\)

đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)

\(< =>t^2-2.2t+4-6^2=0\)

\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)

\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)

a) ta có: 3x + 5 chia hết cho x + 1 

=> 3x + 3 + 2 chia hết cho x + 1 

3.(x+1) + 2 chia hết cho x + 1 

mà 3.(x+1) chia hết cho x + 1 

=> 2 chia hết cho x + 1 

...

bn tự làm tiếp nha! phần b làm tương tự

c) ta có: 2x² + 5x + 7 chia hết cho x -1

=> 2x² -2x + 7x - 7 + 14 chia hết cho x -1

2x.(x-1) + 7.(x-1) + 14 chia hết cho x -1

(x-1).(2x+7) + 14 chia hết cho x -1

mà (x-1).(2x+7) chia hết cho x -1

=> 14 chia hết cho x -1

...

a, 2x-1 thuộc ước của 2,rồi giải ra  

b,c tương tự

d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d

mik cảm ơn nhiều nhé mik cx vừa lam ra ạ

a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

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

b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)

e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)

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

a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)

e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

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