Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. Giải phương trình, hệ phương trình:
a) 2x2 - 5x + 3 = 0
\(\Leftrightarrow2x^2-2x-3x+3=0\)
\(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) x2 - 3x = 0
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\left(x+1\right)-5\left(y+1\right)=5\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left(x+1\right)-15\left(y+1\right)=15\\6\left(x+1\right)-4\left(y+1\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-11\left(y+1\right)=13\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=\dfrac{-13}{11}\\3\left(x+1\right)-2.\left(-\dfrac{13}{11}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\3\left(x+1\right)=-\dfrac{15}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\x=-\dfrac{16}{11}\end{matrix}\right.\)
Hix ,mệt quá.
\(d,\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{163}{y}=-489\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\dfrac{60}{x}+405=525\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) (x – 3)2 + (x + 4)2 = 23 – 3x ⇔ x2 – 6x + 9 + x2 + 8x + 16 = 23 – 3x
⇔ 2x2 + 5x + 2 = 0
∆ = 25 – 16 = 9
x1 = -2, x2 =
b) x3 + 2x2 – (x – 3)2 = (x – 1)(x2 – 2)
⇔ x3 + 2x2 – x2 + 6x – 9 = x3 – x2 – 2x + 2 ⇔ 2x2 + 8x – 11 = 0
∆’ = 16 + 22 = 38
x1 = , x2 =
c) (x – 1)3 + 0,5x2 = x(x2 + 1,5)
⇔ x3 – 3x2 + 3x – 1 + 0,5x2 = x3 + 1,5x
⇔ 2,5x2 – 1,5x + 1 = 0
⇔ 5x2 – 3x + 2 = 0; ∆ = 9 – 40 = -31 < 0
Phương trình vô nghiệm
d) – 1 =
-
⇔ 2x(x – 7) – 6 = 3x – 2(x – 4)
⇔ 2x2 – 14x – 6 = 3x – 2x + 8
⇔ 2x2 – 15x – 14 = 0; ∆ = 225 + 112 = 337
x1 = , x2 =
e) = 1 -
. Điều kiện: x ≠ ±3
Phương trình được viết lại: = 1 +
⇔ 14 = x2 – 9 + x + 3
⇔ x2 + x – 20 = 0, ∆ = 1 + 4 . 20 = 81
√∆ = 9
Nên x1 = = -5; x2 =
= 4 (thỏa mãn)
Vậy phương trình có hai nghiệm x1 = -5, x2 = 4.
f) =
. Điều kiện: x ≠ -1, x ≠ 4
Phương trình tương đương với:
2x(x – 4) = x2 – x + 8 ⇔ 2x2 – 8x – x2 + x – 8 = 0
⇔ x2 – 7x – 8 = 0
Có a – b + c = 1 – (-7) – 8 = 0 nên x1 = -1, x2 = 8
Vì x1 = -1 không thỏa mãn điều kiện của ẩn nên: phương trình có một nghiệm là x = 8.
a) + 2 = x(1 - x)
⇔ x2 – 9 + 6 = 3x – 3x2
⇔ 4x2 – 3x – 3 = 0; ∆ = 57
x1 = , x2 =
b) + 3 =
. Điều kiện x ≠ 2, x ≠ 5.
(x + 2)(2 – x) + 3(x – 5)(2 – x) = 6(x – 5)
⇔ 4 – x2 – 3x2 + 21x – 30 = 6x – 30 ⇔ 4x2 – 15x – 4 = 0
∆ = 225 + 64 = 289, √∆ = 17
x1 = , x2 = 4
c) =
. Điều kiện: x ≠ -1; x ≠ -2
Phương trình tương đương: 4(x + 2) = -x2 – x + 2
⇔ 4x + 8 = 2 – x2 – x
⇔ x2 + 5x + 6 = 0
Giải ra ta được: x1 = -2 không thỏa mãn điều kiện của ẩn nên phương trình chỉ có một nghiệm x = -3.
a) + 2 = x(1 - x)
⇔ x2 – 9 + 6 = 3x – 3x2
⇔ 4x2 – 3x – 3 = 0; ∆ = 57
x1 = , x2 =
b) + 3 =
. Điều kiện x ≠ 2, x ≠ 5.
(x + 2)(2 – x) + 3(x – 5)(2 – x) = 6(x – 5)
⇔ 4 – x2 – 3x2 + 21x – 30 = 6x – 30 ⇔ 4x2 – 15x – 4 = 0
∆ = 225 + 64 = 289, √∆ = 17
x1 = , x2 = 4
c) =
. Điều kiện: x ≠ -1; x ≠ -2
Phương trình tương đương: 4(x + 2) = -x2 – x + 2
⇔ 4x + 8 = 2 – x2 – x
⇔ x2 + 5x + 6 = 0
Giải ra ta được: x1 = -2 không thỏa mãn điều kiện của ẩn nên phương trình chỉ có một nghiệm x = -3.
nhớ like nha
Ta có : \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{3}{10}\)
\(\Leftrightarrow10\left(x+3\right)-10x=3x\left(x+3\right)\)
\(\Leftrightarrow-3x^2-9x+30=0\)
\(\Delta=\left(-9\right)^2+4.3.30=81+360=441>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{9+\sqrt{441}}{-6}=-5\\x_2=\dfrac{9-\sqrt{441}}{-6}=2\end{matrix}\right.\)
Vậy \(S=\left\{-5;2\right\}\)
a) \(\dfrac{12}{x-1}-\dfrac{8}{x+1}=1\) \(\Leftrightarrow\) \(\dfrac{12\left(x+1\right)-8\left(x-1\right)}{x^2-1}=1\)
\(\Leftrightarrow\) \(\dfrac{12x+12-8x+8}{x^2-1}=1\) \(\Leftrightarrow\) \(\dfrac{4x+20}{x^2-1}=1\)
\(\Leftrightarrow\) \(x^2-1=4x+20\) \(\Leftrightarrow\) \(x^2-4x-21=0\)
giải pt ta có 2 nghiệm : \(x_1=7;x_2=-3\)
vậy phương trình có 2 nghiệm \(x=7;x=-3\)
b) \(\dfrac{16}{x-3}+\dfrac{30}{1-x}=3\) \(\Leftrightarrow\) \(\dfrac{16\left(1-x\right)+30\left(x-3\right)}{\left(x-3\right)\left(1-x\right)}=3\)
\(\Leftrightarrow\) \(\dfrac{16-16x+30x-90}{x-x^2-3+3x}=3\) \(\Leftrightarrow\) \(\dfrac{14x-74}{-x^2+4x-3}=3\)
\(\Leftrightarrow\) \(3\left(-x^2+4x-3\right)=14x-74\)
\(\Leftrightarrow\) \(-3x^2+12x-9=14x-74\)
\(\Leftrightarrow\) \(3x^2-2x-65=0\)
giải pt ta có 2 nghiệm : \(x_1=5;x_2=\dfrac{-13}{3}\)
vậy phương trình có 2 nghiệm \(x=5;x=\dfrac{-13}{3}\)
c) ĐK: x\(\ne3,x\ne-2\)
\(\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{1}{x-3}\Leftrightarrow\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}\Leftrightarrow x^2-3x+5=x+2\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy S={1}
d) ĐK: \(x\ne2,x\ne-4\)
\(\dfrac{2x}{x-2}-\dfrac{x}{x+4}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x\left(x+4\right)}{\left(x-2\right)\left(x+4\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x}{\left(x-2\right)\left(x+4\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x-x^2+2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow x^2+10x=8x+8\Leftrightarrow x^2+2x-8=0\Leftrightarrow x^2-2x+4x-8=0\Leftrightarrow x\left(x-2\right)+4\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-4\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
Lời giải:
ĐKXĐ: \(x\neq \pm 1\)
Ta có: \(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\)
\(\Leftrightarrow \left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x-1}.\frac{x}{x+1}=\frac{10}{9}+\frac{2x^2}{(x-1)(x+1)}\)
\(\Leftrightarrow \left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2=\frac{10}{9}+\frac{2x^2}{x^2-1}\)
\(\Leftrightarrow \left(\frac{x(x+1)+x(x-1)}{x^2-1}\right)^2=\frac{10}{9}+\frac{2x^2}{x^2-1}\)
\(\Leftrightarrow \left(\frac{2x^2}{x^2-1}\right)^2=\frac{10}{9}+\frac{2x^2}{x^2-1}\)
Đặt \(\frac{2x^2}{x^2-1}=t\Rightarrow t^2=\frac{10}{9}+t\)
\(\Leftrightarrow 9t^2-9t-10=0\)
\(\Leftrightarrow (3t-5)(3t+2)=0\) \(\Leftrightarrow \left[\begin{matrix} t=\frac{5}{3}\\ t=\frac{-2}{3}\end{matrix}\right.\)
Nếu \(t=\frac{5}{3}\Rightarrow \frac{2x^2}{x^2-1}=\frac{5}{3}\Leftrightarrow 6x^2=5x^2-5\)
\(\Leftrightarrow x^2=-5\) (VL)
Nếu \(t=\frac{-2}{3}\Rightarrow \frac{2x^2}{x^2-1}=\frac{-2}{3}\)
\(\Leftrightarrow 6x^2=2-2x^2\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)(t/m)
Vậy..........