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Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
A) Ta có: \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)
\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)
\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)
\(\Leftrightarrow4x^2-x^2-3x^2+32x-14x-6x=-24+80+40\)
\(\Leftrightarrow12x=96\)
\(\Leftrightarrow x=8\)
Vậy x = 8
B) Ta có: \(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\left(x+2\right)^2-2.8\left(2x+1\right)=25.8+\left(x-2\right)^2\)
\(\Leftrightarrow x^2+4x+4-32x-16=200+x^2-4x+4\)
\(\Leftrightarrow x^2-x^2+4x-32x+4x=200+4-4+16\)
\(\Leftrightarrow-24x=216\)
\(\Leftrightarrow x=-9\)
Vậy x = -9
a)(x+1)(x+2)(x+3)(x+4)+1
=(x+1)(x+4)(x+2)(x+3)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4) thì (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
Thay: =(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16) thì (x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
Thay: =(x2+10x+16+4)2
=(x2+5x+20)2
a)(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4)
Ta có: (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
=(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16)
Ta có:(x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
=(x2+10x+16+4)2
=(x2+5x+20)2
Mk yêu bé Shin-Conan lém
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)
Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)
=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)
=> \(x^2-4x-2x+8-x-2=-2x\)
=> \(x^2-5x+6=0\)
=> \(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)
=> x = 3 .
Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)
b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)
Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)
=> \(x\left(x+12\right)=192\)
=> \(x^2+12x-192=0\)
=> \(x^2+2x.6+36-228=0\)
=> \(\left(x+6\right)^2=288\)
=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)
tôi chịu
b) Đặt \(x-7=a\) ta có:
\(\left(a+1\right)^4+\left(a-1\right)^4=16\)
\(\Leftrightarrow\)\(a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16\)
\(\Leftrightarrow\)\(2a^4+12a^2+2-16=0\)
\(\Leftrightarrow\)\(2\left(a^4+6a^2-7\right)=0\)
\(\Leftrightarrow\)\(a^4+6a^2-7=0\)
\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
Vì \(a^2+7>0\) nên \(\orbr{\begin{cases}a-1=0\\a+1=0\end{cases}}\)
Thay trở lại ta có: \(\orbr{\begin{cases}x-8=0\\x-6=0\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
Vậy...