Giải phương trình
a) \((x^2-1)(x+2)(x-3)=(x-1)(x^2-4)(x+5)\)
b) \( (5x2 - 2x + 10)^2 = (x^2 + 10x - 8)^2\)
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Những câu hỏi liên quan
10 tháng 7 2021
a) Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{x-1}+1+1\)(Vô lý)
Vậy: \(S=\varnothing\)
b) Ta có: \(\sqrt{x^4+2x^2+1}=\sqrt{x^2+10x+25}-10x+22\)
\(\Leftrightarrow x^2+1=\left|x+5\right|-10x+22\)
\(\Leftrightarrow\left|x+5\right|=x^2+1+10x-22=x^2+10x-21\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+10x-21\left(x\ge-5\right)\\-x-5=x^2+10x-21\left(x< -5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x-21-x-5=0\\x^2+10x-21+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+9x-26=0\\x^2+11x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{185}}{2}\\x=\dfrac{-11-\sqrt{185}}{2}\end{matrix}\right.\)
BC
2
LT
6 tháng 3 2021
a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow20-4x=0\)
\(\Leftrightarrow4x=20\)
\(\Leftrightarrow x=5\)
Vậy S = {5}
b) ĐKXĐ: \(x\ne0;x\ne-2\)
\(\dfrac{x+2}{x}=\dfrac{\left(x+1\right)\left(x+4\right)}{x^2+2x}+\dfrac{x}{x+2}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)
\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)
\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)
\(\Leftrightarrow x^3+x^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy S = {-1}
c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!
6 tháng 3 2021
a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4=x^2-16\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow-4x+20=0\)
\(\Leftrightarrow-4x=-20\)
hay x=5
Vậy: S={5}
8 tháng 11 2023
a:
ĐKXĐ: \(x>=-2\)
\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)
=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)
Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)
Phương trình sẽ trở thành:
1+ab=a+b
=>ab-a-b+1=0
=>a(b-1)-(b-1)=0
=>(b-1)(a-1)=0
=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)
=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)
=>\(x\in\varnothing\)
b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)
TH1: x>=1/4
\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)
=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)
=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)
=>4x-1=0
=>x=1/4(nhận)
TH2: x<1/4
Phương trình (1) sẽ trở thành:
\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)
=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)
=>4x-1=0
=>x=1/4(loại)
8T
1
18 tháng 1 2022
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
12 tháng 2 2022
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
KK
9 tháng 5 2021
a. \(\dfrac{-3}{x^2-9}+\dfrac{5}{3-x}=\dfrac{2}{x+3}\)
<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5}{x-3}=\dfrac{2}{x+3}\)
<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5\left(x+3\right)}{x^2-9}=\dfrac{2\left(x-3\right)}{x^2-9}\)
<=> \(-3+\left(-5\right)\left(x+3\right)=2\left(x-3\right)\)
<=> -3 + (-5x) + (-15) = 2x - 6
<=> -5x -2x = 15 - 6 + 3
<=> -7x = 12
<=> x = \(\dfrac{-12}{7}\)
Vậy ........
b. \(\left|x+5\right|=2x-1\)
Nếu x \(\ge\) -5 => \(\left|x+5\right|\) = x + 5
Nếu x < -5 => \(\left|x+5\right|\) = -(x + 5)
TH1: Nếu x \(\ge\) -5
<=> x + 5 = 2x - 1
<=> x - 2x = -1 - 5
<=> -x = -6
<=> x = 6
TH2: Nếu x < -5
<=> -(x + 5) = 2x - 1
<=> -x - 5 = 2x - 1
<=> -5 + 1 = 2x + x
<=> -4 = 3x
<=> x = \(\dfrac{-4}{3}\)
Vậy .........
c. Bạn tự giải câu này nhé (có thể tách các hạng tử rồi tính)
NN
9 tháng 6 2021
a) \(2\chi-3=3\left(\chi+1\right)\)
\(\Leftrightarrow2\chi-3=3\chi+3\)
\(\Leftrightarrow2\chi-3\chi=3+3\)
\(\Leftrightarrow\chi=-6\)
Vậy phương trình có tập nghiệm S= \(\left\{-6\right\}\)
\(3\chi-3=2\left(\chi+1\right)\)
\(\Leftrightarrow3\chi-3=2\chi+2\)
\(\Leftrightarrow3\chi-2\chi=2+3\)
\(\Leftrightarrow\chi=5\)
Vậy phương trình có tập nghiệm S= \(\left\{5\right\}\)
b) \(\left(3\chi+2\right)\left(4\chi-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi+2=0\\4\chi-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-2\\4\chi=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-2}{3}\\\chi=\dfrac{5}{4}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-2}{3};\dfrac{5}{4}\right\}\)
\(\left(3\chi+5\right)\left(4\chi-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi+5=0\\4\chi-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-5\\4\chi=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-5}{3}\\\chi=\dfrac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-5}{3};\dfrac{1}{2}\right\}\)
c) \(\left|\chi-7\right|=2\chi+3\)
Trường hợp 1:
Nếu \(\chi-7\ge0\Leftrightarrow\chi\ge7\)
Khi đó:\(\left|\chi-7\right|=2\chi+3\)
\(\Leftrightarrow\chi-7=2\chi+3\)
\(\Leftrightarrow\chi-2\chi=3+7\)
\(\Leftrightarrow\chi=-10\) (KTMĐK)
Trường hợp 2:
Nếu \(\chi-7\le0\Leftrightarrow\chi\le7\)
Khi đó: \(\left|\chi-7\right|=2\chi+3\)
\(\Leftrightarrow-\chi+7=2\chi+3\)
\(\Leftrightarrow-\chi-2\chi=3-7\)
\(\Leftrightarrow-3\chi=-4\)
\(\Leftrightarrow\chi=\dfrac{4}{3}\)(TMĐK)
Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{4}{3}\right\}\)
\(\left|\chi-4\right|=5-3\chi\)
Trường hợp 1:
Nếu \(\chi-4\ge0\Leftrightarrow\chi\ge4\)
Khi đó: \(\left|\chi-4\right|=5-3\chi\)
\(\Leftrightarrow\chi-4=5-3\chi\)
\(\Leftrightarrow\chi+3\chi=5+4\)
\(\Leftrightarrow4\chi=9\)
\(\Leftrightarrow\chi=\dfrac{9}{4}\)(KTMĐK)
Trường hợp 2: Nếu \(\chi-4\le0\Leftrightarrow\chi\le4\)
Khi đó: \(\left|\chi-4\right|=5-3\chi\)
\(\Leftrightarrow-\chi+4=5-3\chi\)
\(\Leftrightarrow-\chi+3\chi=5-4\)
\(\Leftrightarrow2\chi=1\)
\(\Leftrightarrow\chi=\dfrac{1}{2}\)(TMĐK)
Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)
1 tháng 11 2023
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
H
1 tháng 11 2023
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+1\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(7-5x\right)=0\)
\(\Leftrightarrow x-1=0;x+2=0\)hoặc \(7-5x=0\)
\(\Leftrightarrow x=1;x=-2\)hoặc \(x=\frac{7}{5}\)
KL....
\(b,\left(5x^2-2x+10\right)^2=\left(x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-x^2-10x+8\right)\left(5x^2-2x+10+x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(4x^2-12x+18\right)\left(6x^2+8x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+\frac{9}{2}\right)\left(6x^2+6x+2x+2\right)=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{9}{4}\right)\left(6x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\right]\left(3x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-1\end{cases}}\)Vì \(\left(x-\frac{3}{2}\right)^2+\frac{9}{4}>0\forall x\)
Vậy ..