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\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)[x\left(x-4\right)-3\left(x-4\right)]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\)
Tương đương với 1 trong 3 biểu thức trên bằng 0.
Giải ra 3 nghiệm là \(x=\frac{1}{3};x=4;x=3\)
a) \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
<=> \(9x^2-9x+2=9x^2+6x+1\)
<=> \(15x=1\) <=> \(x=\frac{1}{15}\)
b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)
<=> \(4x^2+3x-1=4x^2-12x+9\)
<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)
c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)
<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)
<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)
d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)
<=> 16 - 9x2 = 12x - 9x2 - 3
<=> 12x = 19
<=> x = 19/12
e) x(x + 1)(x + 2)(x + 3) = 24
<=> (x2 + 3x)(x2 + 3x + 2) = 24
<=> (x2 + 3x)2 + 2(x2 + 3x) - 24 = 0
<=> (x2 + 3x)2 + 6(x2 + 3x) - 4(x2 + 3x) - 24 = 0
<=> (x2 + 3x + 6)(x2 + 3x - 4) = 0
<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
g) (7x - 2)2 = (7x - 3)(7x + 2)
<=> 49x2 - 28x + 4 = 49x2 - 7x - 6
<=> 21x = 10 <=> x = 10/21
\(a,\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
\(9x^2-3x-6x+2=9x^2+6x+1\)
\(-9x+2-6x-1=0\)
\(-15x+1=0\)
\(-15x=-1\)
\(x=\frac{1}{15}\)
giải phương trình bất nhất (3x-1)(x+3)= (2-x)(5-3x) các bạn ghi các bước giải ra giúp mik luôn nha !
(3x-1)(x+3)= (2-x)(5-3x)
\(\Leftrightarrow3x^2+9x-x-3=10-6x-5x+3x^2\)
\(\Leftrightarrow3x^2+8x-3-10+11x-3x^2=0\)
\(\Leftrightarrow19x-13=0\)
\(\Leftrightarrow x=\frac{13}{19}\)
Vậy \(x\in\left\{\frac{13}{19}\right\}\)
a) \(2x^3 + 6x^2 = x^2 +3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\)
\(\Leftrightarrow\left(x+3\right).x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
S = \(\left\{-3;0;\dfrac{1}{2}\right\}\)
b) \((3x-1) (x^2 +2 ) = (3x-1) (7x - 10)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
S = \(\left\{\dfrac{1}{3};3;4\right\}\)
\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)
Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
a,xet khoang -5/3<x<1/3 suy ra 2x+5=3x-1 suy ra x=6 khong thoa man
xet khoang x<-5/3 suy ra -2x-5=3x-1 suy ra x=-4/5 tm
xet khong x>1/3 suy ra x=-4/5 ktm
vay.............................
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
<=> \(\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
<=> \(\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
<=> 3x -1 = 0 hoặc x - 3 = 0 hoặc x - 4 = 0
<=> x = 1/3 hoặc x = 3 hoặc x = 4
Vậy S = { 1/3 ; 3; 4 }