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\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\)
\(\Leftrightarrow2x-3x^2-x< 15-3x^2-6x\)
\(\Leftrightarrow x-3x^2+3x^2+6x< 15\)
\(\Leftrightarrow7x< 15\)
\(\Leftrightarrow x< \frac{15}{7}\)
\(b,\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)
\(\Leftrightarrow\frac{1-2x}{4}-\frac{1-5x}{8}-x\le2\)
\(\Leftrightarrow\frac{2-4x}{8}-\frac{1-5x}{8}-\frac{8x}{8}\le2\)
\(\Leftrightarrow\frac{2-4x-1+5x-8x}{8}\le2\)
\(\Leftrightarrow-7x+1\le16\)
\(\Leftrightarrow-7x\le15\)
\(\Leftrightarrow x\le-\frac{15}{7}\)
\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\)
\(2x-3x^2-x< 15-3x^2-6x\)
\(2x-x+6x-3x^2+3x^2< 15\)
\(7x< 15\)
\(x< \frac{15}{7}\)
\(b,\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)
\(2\left(1-2x\right)-16\le1-5x+8x\)
\(2-4x-16\le1+3x\)
\(-14-4x\le1+3x\)
\(-4x-3x\le1+14\)
\(-7x\le15\)
\(x\ge-\frac{15}{7}\)
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\)\(\left(3x-2\right)\left(3x-6\right)\)
\(=\)\(3\left(x-2\right)\left(3x-2\right)\)
Chúc bạn học tốt ~
a)(2x+1)(3x-2)=(5x-8)(2x+1)
⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0
⇔(2x+1)(3x-2-5x+8)=0
⇔(2x+1)(-2x+6)=0
⇔2x+1=0 hoặc -2x+6=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-2x+6=0⇔-2x=-6⇔x=3
phương trình có 2 nghiệm x=-1/2 và x=3
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(3x-2\right)^2-\left(2x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(x-2\right)\left(5x-2\right)=0\end{matrix}\right.\)
hay \(x\in\left\{2;\dfrac{2}{5}\right\}\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x< =0\\\left(2x+4\right)^2-\left(4x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =0\\\left(-2x+4\right)\left(6x+4\right)=0\end{matrix}\right.\)
hay x=-2/3
c: \(\Leftrightarrow\left\{{}\begin{matrix}x< =21\\\left(2x-3\right)^2-\left(x-21\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =21\\\left(2x-3-x+21\right)\left(2x-3+x-21\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =21\\\left(x+18\right)\left(3x-24\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-18;8\right\}\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-1-x+2\right)\left(3x-1+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x+1\right)\left(4x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)
<=> \(\left(3x-2\right)\left(x+1\right)^2.3^2.\left(3x+8\right)+144=0\)
<=> \(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\) (*)
Đặt \(3x+3=t\) Khi đó pt (*) trở thành:
\(\left(t-5\right)t^2\left(t+5\right)+144=0\)
<=> \(t^4-25t^2+144=0\)
<=> \(\left(t-4\right)\left(t-3\right)\left(t+3\right)\left(t+4\right)=0\)
đến đây bn tự giải tiếp nhé
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)
\(\Rightarrow\left(3x+2+3x-2\right)\left(3x+2-3x+2\right)=5x+8\)
\(\Rightarrow4.6x=5x+8\Rightarrow24x=5x+8\)
\(\Rightarrow19x=8\Rightarrow x=\frac{8}{19}\)
b) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Rightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Rightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
\(\Rightarrow-12x+12+9x-9=3x-9\)
\(\Rightarrow-3x+3=3x-9\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\left(x^2-3x+1\right)\left(21+3x-x^2\right)=121\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left[22-\left(x^2-3x+1\right)\right]=121\)
Đặt \(x^2-3x+1=a\)
\(\Rightarrow a\left(22-a\right)=121\)
\(\Leftrightarrow a^2-22a+121=0\)
\(\Leftrightarrow\left(a-11\right)^2=0\)
\(\Leftrightarrow a=11\)
\(\Rightarrow x^2-3x+1=11\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-2;5\right\}\)
a ) Đặt t = x - 7 .
=> pt trở thành : \(\left(t-1\right)^4+\left(t+1\right)^4=16\)
b ) Phân tích :
\(x^3-x^2+2x^2-2x+3x-3=0\)
NHhóm rồi sẽ ra
Nhã Doanh, đề bài khó wá, TNA Atula, Đinh Đức Hùng, Gia Hân Ngô, Phương Ann, Vũ Elsa, Aki Tsuki, Mysterious Person, Nguyễn Thị Bích Thủy, nguyen thi vang, Mashiro Shiina, Hung nguyen, ...
Tks p!!! Phương Ann
Giúp mk thêm 2 PT này nha!!!
a, \(\left(x-6\right)^4+\left(x-8\right)^4=16\)
b, \(x^3+x^2+x-3=0\)
Phương Ann