giai cac pt
d.(3x-1)(x^2+2)=(3x-1)(7x-10)
K
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25 tháng 1 2021
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
17 tháng 1 2018
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy tập nghiệm của ohuowng trình là \(S=\left\{\dfrac{1}{3};3;4\right\}\)
LT
1
10 tháng 7 2023
=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)
=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)
=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)
=>2x^2-10x+8=3
=>2x^2-10x+5=0
=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2020
Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:
ĐKXĐ: ...
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)
Đặt \(4x+\frac{7}{x}-10=t\)
\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)
DV
2
TA
30 tháng 1 2018
a) 2x2-4x-x+2=0
=> 2x(x-2)-(x-2)=0
=> (2x-1)(x-2)=0
=> \(\left[{}\begin{matrix}2x-1=0\\x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
b) 3x2-12x+5x-20=0
=> 3x(x-4)+5.(x-4)=0
=> (x-4)(3x+5)=0
=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)
c)x3+2x2-x2-2x+2x+4=0
=> x2(x+2)-x(x+2)+2(x+2)=0
=>(x2-x+2)(x+2)=0
=> x=-2( vi x2-x+2>0)
d) x3-x2-4x2+4x+4x-4=0
=> x2(x-1)-4x(x-1)+4(x-1)=0
=>(x-1)(x2-4x+4)=0
=> \(\left[{}\begin{matrix}x-1=0\\x^2-4x+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
KK
30 tháng 1 2018
2x2-5x+2=0
⇔2x2-x-4x+2=0
⇔x(2x-1)-2(2x-1)=0
⇔(x-2)(2x-1)=0
⇔\(\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\2x=1\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)
sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)
x3+x2+4=0
⇔x3+2x2-x2-2x+2x+4=0
⇔(x3+2x2)-(x2+2x)+(2x+4)=0
⇔x2(x+2)-x(x+2)+2(x+2)=0
⇔(x+2)(x2-x+2)=0
⇔x+2=0 và x2-x+2=0
⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)
vậy S={-2}
AH
Akai Haruma
Giáo viên
13 tháng 10 2018
Lời giải:
ĐKXĐ: \(x\leq \frac{2}{3}\)
Ta có: \(\sqrt{2-3x}=-3x^2+7x-1\)
\(\Leftrightarrow 3x^2-7x+1+\sqrt{2-3x}=0\)
\(\Leftrightarrow x(3x-1)-2(3x-1)+\sqrt{2-3x}-1=0\)
\(\Leftrightarrow x(3x-1)-2(3x-1)+\frac{2-3x-1}{\sqrt{2-3x}+1}=0\)
\(\Leftrightarrow (3x-1)\left(x-2-\frac{1}{\sqrt{2-3x}+1}\right)=0\)
Vì \(x\leq \frac{2}{3}; \frac{1}{\sqrt{2-3x}+1}>0\Rightarrow x-2-\frac{1}{\sqrt{2-3x}+1}< \frac{2}{3}-2-0<0\)
Tức là \(x-2-\frac{1}{\sqrt{2-3x}+1}\neq 0\Rightarrow 3x-1=0\Rightarrow x=\frac{1}{3}\) (t/m)
Vậy...........
MD
1
17 tháng 2 2021
a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)
b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)
c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)
15 tháng 2 2017
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
KT
1
HP
6 tháng 1 2021
ĐK: \(x\ge1\)
Đặt \(\sqrt{3x-2}+2\sqrt{x-1}=t\left(t\ge1\right)\)
\(pt\Leftrightarrow3t=t^2-4\)
\(\Leftrightarrow t^2-3t-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-1\left(l\right)\end{matrix}\right.\)
\(t=4\Leftrightarrow\sqrt{3x-2}+2\sqrt{x-1}=4\)
\(\Leftrightarrow7x-6+4\sqrt{\left(3x-2\right)\left(x-1\right)}=16\)
\(\Leftrightarrow4\sqrt{3x^2-5x+2}=22-7x\)
\(\Leftrightarrow\left\{{}\begin{matrix}48x^2-80x+32=484+49x^2-308x\\22-7x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}452+x^2-228x=0\\x\le\dfrac{22}{7}\end{matrix}\right.\)
\(\Leftrightarrow x=2\left(tm\right)\)
(3x-1) (x2 +2) = (3x-1)(7x-10)
=> (3x-1) (x2+2)-(3x-1)(7x-10)=0
=>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x2-7x+12)=0
=>(3x-1)(x-3)(x-4)=0
=>3x-1=0 => x= 1/3
x-3=0 => x=3
x-4=0 => x=4
vậy pt có tập nghiệm S={ 1/3; 3; 4}
mk lam xong roi ban moi giai