Giải phương trình \(\left(3x-1\right)\left(x+2\right)=\left(3x-1\right)\)\(\left(7x-10\right)\)
Tập nghiệm của phương trình là:
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( 3x - 1)( x + 2) = ( 3x - 1)(7x - 10)
<=>( 3x - 1)( x + 2) - ( 3x - 1)(7x - 10) = 0
<=> ( 3x - 1)( x + 2 - 7x + 10) = 0
<=>( 3x - 1)( -6x + 12) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\-6x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}}\)
Vậy.....
\(\left(3x-1\right)\left(x+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(3x^2+5x-2=21x^2-37x+10\)
\(3x^2+5x-2-21x^2+37x-10=0\)
\(-18x^2+42x-12=0\)
\(-6\left(3x-1\right)\left(x-2\right)=0\)
\(-6\ne0\)
\(\left(3x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=1\\x=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}}\)
a) Phương trình: \({x^2} - 3x + 2 = 0\,\,\,\left( 1 \right)\)
Ta có: \(\Delta = 9 - 4.2 = 1 > 0\)
Phương trình (1) có hai nghiệm \(\left\{ \begin{array}{l}{x_1} = \frac{{3 + 1}}{{2.1}} = 2\\{x_1} = \frac{{3 - 1}}{{2.1}} = 1\end{array} \right.\) => \({S_1} = \left\{ {1;2} \right\}\)
Phương trình: \(\left( {x - 1} \right)\left( {x - 2} \right) = 0\,\,\,\left( 2 \right)\)\( \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 2\end{array} \right.\) => \({S_2} = \left\{ {1;2} \right\}\)
b) Hai tập \({S_1};{S_2}\) có bằng nhau
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 = \(\left\{\dfrac{1}{3};3;4\right\}\)
\(-4x+7=-1\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)
Vậy phương trình có tập nghiệm \(S=\left\{2\right\}\)
\(\frac{\left(3x+2\right)\left(x+2\right)}{2}-\frac{3}{2}\left(x+1\right)^2=\frac{x-1}{2}\)
\(\Leftrightarrow3x^2+2x+6x+4-3\left(x^2+2x+1\right)=x-1\)
\(\Leftrightarrow3x^2+2x+6x+4-3x^2-6x-3-x+1=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy pt đã cho có nghiệm \(x=-2\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
c.
\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)
\(\Leftrightarrow2cos\left(x+12^0\right)=1\)
\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)
2.
Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:
\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)
\(\Rightarrow-1\le m\le\dfrac{1}{2}\)
a.
\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)
b.
\(2x-10^0=arccot\left(4\right)+k180^0\)
\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)
\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)
Lời giải:
a. $f'(x)\leq 0$
$\Leftrightarrow 3x^2-6x\leq 0$
$\Leftrightarrow x(x-2)\leq 0$
$\Leftrightarrow 0\leq x\leq 2$
b.
$f'(x)=x^2-3x+2=0$
$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$
$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
c.
$g(x)=f(1-2x)+x^2-x+2022$
$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$
$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$
$g'(x)\geq 0$
$\Leftrightarrow -24x^2+2x+5\geq 0$
$\Leftrightarrow (5-12x)(2x-1)\geq 0$
$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$
1, <=> 3x-1-2x=6 <=> x =7
2, (2x-1)(7-x)=x^2-7x
<=> 14x -2x^2-7+x=x^2-7x
<=> -3x^2+ 15x - 7 = -7x
<=> -3x^2 +23x - 7 =0
<=> \(x=\dfrac{23\pm\sqrt{445}}{6}\)
1.
\(\Leftrightarrow3x-1-2x=6\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=7\)
2.
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+7x-x^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+x\left(7-x\right)=0\)
\(\Leftrightarrow\left(7-x\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(\sqrt{2-f\left(x\right)}=f\left(x\right)\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)\ge0\\f^2\left(x\right)+f\left(x\right)-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\f\left(x\right)=-2< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow f\left(1\right)=f\left(2\right)=f\left(3\right)=1\)
\(\sqrt{2g\left(x\right)-1}+\sqrt[3]{3g\left(x\right)-2}=2.g\left(x\right)\)
\(VT=1.\sqrt{2g\left(x\right)-1}+1.1\sqrt[3]{3g\left(x\right)-2}\)
\(VT\le\dfrac{1}{2}\left(1+2g\left(x\right)-1\right)+\dfrac{1}{3}\left(1+1+3g\left(x\right)-2\right)\)
\(\Leftrightarrow VT\le2g\left(x\right)\)
Dấu "=" xảy ra khi và chỉ khi \(g\left(x\right)=1\)
\(\Rightarrow g\left(0\right)=g\left(3\right)=g\left(4\right)=g\left(5\right)=1\)
Để các căn thức xác định \(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)-1\ge0\\g\left(x\right)-1\ge0\end{matrix}\right.\)
Ta có:
\(\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+f\left(x\right).g\left(x\right)-f\left(x\right)-g\left(x\right)+1=0\)
\(\Leftrightarrow\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+\left[f\left(x\right)-1\right]\left[g\left(x\right)-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\g\left(x\right)=1\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy tập nghiệm của pt đã cho có đúng 1 phần tử
Bạn bạn nhân phân phối (3x-1)(x-2) và (3x-1)(7x-10)
Sau đó chuyển vế sao cho về phương trình bậc 2
Sau đó giải pt bậc hai là ra
Ta có : (3x -1 ) . ( x + 2 ) = ( 3x-1 ) .( 7x - 10)
<=>3.x2 + 6x -x -2 = 21x2 -30x - 7x +10
<=> 3x2 + 5x -2 = 21x2 -37x + 10
<=> 3x2 +5x - 3 - 21x2 +37x -10 = 0
<=> -18x2 + 42x -12 = 0
<=> 3x2 -7x +2 = 0
<=> 3x2 -x -6x + 2 = 0
<=> x. ( 3x -1 ) -2.(3x -1 ) = 0
<=> (3x -1 ) . ( x - 2 ) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
Tập nghiệm của phương trình là : { \(\frac{1}{3}\); 2}