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\(a,x^3-x^2-12x+45=0\)
\(\left(x-3\right)\left(x-3\right)\left(x+5\right)=0\)
\(x=3;3;-5\)
\(b,2x^3-5x^2+8x-5=0\)
\(\left(2x^2-3x+5\right)\left(x-1\right)=0\)
\(x=1\)
lm 1 câu đã chán ngắt , giải mấy câu nữa não tớ nổ bùmmm , tớ bt đây là trang web để hc nhưng tạo nên tiếng cười là chính nha ^^
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
a) \(\frac{x+\frac{x+1}{5}}{3}=1-\frac{2x-\frac{1-2x}{34}}{5}\)
\(\Leftrightarrow\frac{\frac{5x+x+1}{5}}{3}=1-\frac{\frac{68x-1+2x}{34}}{5}\)
\(\Leftrightarrow\frac{6x+1}{15}=1-\frac{70-1}{170}\)
\(\Leftrightarrow\frac{6x+1}{15}+\frac{70x-1}{170}-1=0\)
\(\Leftrightarrow\frac{34\left(6x+1\right)+3\left(70x-1\right)-510}{510}=0\)
\(\Leftrightarrow204x+34+210x-3-510=0\)
\(\Leftrightarrow414x-479=0\)
\(\Leftrightarrow x=\frac{479}{414}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{479}{414}\right\}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
a/ \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
<=> \(\frac{\left(x+1\right)^2}{\left(x+1\right)^2+1}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2+2}=\frac{7}{6}\left(1\right)\)
đặt \(\left(x+1\right)^2=a\left(a>0\right)\)
=> \(\left(1\right)\)<=> \(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
<=> \(\frac{a\left(a+2\right)+\left(a+1\right)^2}{\left(a+1\right)\left(a+2\right)}=\frac{7}{6}\)
<=> \(\frac{2a^2+4a+1}{a^2+3a+2}=\frac{7}{6}\)
<=> \(6\left(2a^2+4a+1\right)=7\left(a^2+3a+2\right)\)
<=> \(5a^2+3a-8=0\)
<=> \(5a^2-5a+8a-8=0\)
<=> \(\left(5a+8\right)\left(a-1\right)=0\)
<=> \(a=\frac{-8}{5}\left(h\right)a=1\)
mà \(a>0\)
=> \(a=1\)
=> \(\left(x+1\right)^2=1\)
=> \(x+1=1\left(h\right)x+1=-1\)
=> \(x=0\left(h\right)x=-2\)
vậy ......
chúc bn học tốt
Xét x = 0 và x = -2 , thay vào ta được \(VT=VP\)
Xét x > 0 :
\(VT=\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=1-\frac{1}{x^2+2x+2}+1-\frac{1}{x^2+2x+3}\)
\(=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)>\frac{7}{6}=VP\) ( loại )
Xét x < -2 :
\(VT=2-\left(\frac{1}{x\left(x+2\right)+2}+\frac{1}{x\left(x+2\right)+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{7}{6}=VP\) ( loại )
Xét -2 < x < 0 :
\(VT=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{-2}+1\right)=\frac{3}{2}>\frac{7}{6}=VP\) ( loại )
Vậy ...
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
\(a.\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\)\(0\)
\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2.\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\sqrt{3}\)
\(b.2x^3-5x^2+3x=0\)
\(\Leftrightarrow x.\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x-1\right).\left(2x-3\right)=0\)
Đến đây tự làm nhé có việc bận
câu a sai dzoii