\(\frac{3}{x^2-x+1}+\frac{11}{x^2+x+2}=\frac{54}{x^2+5x+4}\)
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BG
1
OM
5 tháng 5 2020
Giải phương trình
a, 5x(x-4)-5x2 = 2 (11-x)
\(\Leftrightarrow5x^2-20x-5x^2=22-2x\)
\(\Leftrightarrow-18x=22\)
\(\Leftrightarrow x=\frac{-22}{18}\)
b, \(\frac{3}{x-3}-\frac{2}{x+3}=\frac{4x}{x^2-9}\left(x\ne\pm3\right)\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4x}{x^2-9}\)
\(\Rightarrow3x+9-2x+6=4x\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\left(tm\right)\)
Kl: a,.........
b,.........
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
1 tháng 6 2019
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
.....
1 tháng 6 2019
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
.....
M
4 tháng 2 2020
a/ \(7x-5=13-5x\)
\(\Leftrightarrow7x+5x=13+5\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\frac{3}{2}\)
b/\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28=19-2x-22\)
\(\Leftrightarrow10x-20x+2x=19-22-28+15\)
\(\Leftrightarrow-8x=-16\)
\(\Leftrightarrow x=2\)
c/ \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\frac{7\left(2x-1\right)-3\left(5x+2\right)-21\left(x+13\right)}{21}=0\)
\(\Leftrightarrow14x-7-15x-6-21x-273=0\)
\(\Leftrightarrow-22x-286=0\)
\(\Leftrightarrow x=-13\)
e/ \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)\left(x+2\right)-\left(x+1\right)\left(x+2\right)-\left(3x-11\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x^2-4\right)-\left(x^2+3x+2\right)-\left(3x^2-17x+22\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow2x^2-8-x^2-3x-2-3x^2+17x-22=0\)
\(\Leftrightarrow-2x^2+14x-32=0\)
\(\Leftrightarrow x^2-7x+16=0\)
\(\Leftrightarrow x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4.1.16}}{2}\)
\(\Leftrightarrow x=\frac{7\pm\sqrt{-15}}{2}\left(ktm\right)\)
\(\Leftrightarrow x\in\varnothing\)
VM
4 tháng 2 2020
Bài 1:
a) \(7x-5=13-5x\)
\(\Leftrightarrow7x+5x=13+5\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=18:12\)
\(\Leftrightarrow x=\frac{3}{2}.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{3}{2}\right\}.\)
b) \(5.\left(2x-3\right)-4.\left(5x-7\right)=19-2.\left(x+11\right)\)
\(\Leftrightarrow10x-15-\left(20x-28\right)=19-\left(2x+22\right)\)
\(\Leftrightarrow10x-15-20x+28=19-2x-22\)
\(\Leftrightarrow13-10x=-3-2x\)
\(\Leftrightarrow13+3=-2x+10x\)
\(\Leftrightarrow16=8x\)
\(\Leftrightarrow x=16:8\)
\(\Leftrightarrow x=2.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)
c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\frac{7.\left(2x-1\right)}{7.3}-\frac{3.\left(5x+2\right)}{3.7}=\frac{21.\left(x+13\right)}{21}\)
\(\Leftrightarrow\frac{14x-7}{21}-\frac{15x+6}{21}=\frac{21x+273}{21}\)
\(\Leftrightarrow14x-7-\left(15x+6\right)=21x+273\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow-x-13=21x+273\)
\(\Leftrightarrow-x-21x=273+13\)
\(\Leftrightarrow-22x=286\)
\(\Leftrightarrow x=286:\left(-22\right)\)
\(\Leftrightarrow x=-13.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-13\right\}.\)
Chúc bạn học tốt!
Lời giải:
ĐK: $x\neq -1; x\neq -4$
PT \(\Leftrightarrow \frac{3}{x^2-x+1}-\frac{27}{x^2+5x+4}+\frac{11}{x^2+x+2}-\frac{27}{x^2+5x+4}=0\)
\(\Leftrightarrow \frac{3(x^2+5x+4)-27(x^2-x+1)}{(x^2-x+1)(x^2+5x+4)}+\frac{11(x^2+5x+4)-27(x^2+x+2)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{3(-8x^2+14x-5)}{(x^2-x+1)(x^2+5x+4)}+\frac{2(-8x^2+14x-5)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{-8x^2+14x-5}{x^2+5x+4}\left(\frac{3}{x^2-x+1}+\frac{2}{x^2+x+2}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ với mọi $x\neq -1; x\neq -4$
Do đó \(\frac{-8x^2+14x-5}{x^2+5x+4}=0\Rightarrow -8x^2+14x-5=0\)
\(\Rightarrow x=\frac{1}{2}\) hoặc $x=\frac{5}{4}$ (đều thỏa mãn)
Vậy........
Lời giải:
ĐK: $x\neq -1; x\neq -4$
PT \(\Leftrightarrow \frac{3}{x^2-x+1}-\frac{27}{x^2+5x+4}+\frac{11}{x^2+x+2}-\frac{27}{x^2+5x+4}=0\)
\(\Leftrightarrow \frac{3(x^2+5x+4)-27(x^2-x+1)}{(x^2-x+1)(x^2+5x+4)}+\frac{11(x^2+5x+4)-27(x^2+x+2)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{3(-8x^2+14x-5)}{(x^2-x+1)(x^2+5x+4)}+\frac{2(-8x^2+14x-5)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{-8x^2+14x-5}{x^2+5x+4}\left(\frac{3}{x^2-x+1}+\frac{2}{x^2+x+2}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ với mọi $x\neq -1; x\neq -4$
Do đó \(\frac{-8x^2+14x-5}{x^2+5x+4}=0\Rightarrow -8x^2+14x-5=0\)
\(\Rightarrow x=\frac{1}{2}\) hoặc $x=\frac{5}{4}$ (đều thỏa mãn)
Vậy........