nghiem cua pt
\(\frac{x+16}{49}\)+\(\frac{x+18}{47}\)=\(\frac{x+20}{45}\)-1
giải phương trình
\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
Thêm 2 vào pt có :
\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\) (1)
\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\) (2)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{41}-1\)
Mình mới học lớp 6 thui . kb voi mink nhé
Bài 7 : Giải phương trình : \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
Giúp mình với , cần gấp
Lời giải:
PT $\Leftrightarrow \frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1$
$\Leftrightarrow \frac{x+65}{49}+\frac{x+65}{47}=\frac{x+65}{45}$
$\Leftrightarrow (x+65)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0$
Thấy rằng $\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\neq 0$
Do đó $x+65=0\Rightarrow x=-65$
a. \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\)
b.\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
c.\(\frac{7x-1}{2}=5+\frac{9-5x}{6}\)
d.\(\frac{3x-9}{x+1}-2=\frac{4x}{x+1}\)
c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)
\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)
\(\Rightarrow42x-6=60+18-10x\)
\(\Leftrightarrow52x-84=0\)
\(\Leftrightarrow x=\dfrac{21}{13}\)
Vậy....
d) tương tự
a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
Vậy....
b) \(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}-1\)
\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}-1+1+1\)
\(\Leftrightarrow\dfrac{x+16+49}{49}+\dfrac{x+18+47}{47}=\dfrac{x+20+45}{45}\)
\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)
\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)
Vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy....
ngiệm của phương trình : \(\frac{x+16}{49}\)+ \(\frac{x+18}{47}\) = \(\frac{x+20}{45}\) - 1
đúng là toán 8 khó thật nhìn mà hoa cả mắt *_* T_T
duyệt đi
chẳng hoa j cả
áp dụng tỉ lệ thức ta có :
\(\Leftrightarrow\frac{96x+1634}{2303}=\frac{x-25}{45}\Rightarrow\left(96x+1634\right)45=2303\left(x-25\right)\)
tự giải tiếp ra
=>x=-65
Ta có \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Mà \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)
\(\Rightarrow x+65=0\)
\(\Rightarrow x=-65\)
Bài này dễ mà bạn để ý một chút là giải ra ngay.
\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{41}-1\) GIẢI PHƯƠNG TRÌNH NÀY GIÚP MÌNH NHA MƠN NHIỀU
Giải phương trình
a,\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
b,\(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
c,(2x-5)3-(3x-4)3+(x+1)3=0
d,(x2+3x-4)3+(3x2+7x+4)3=(4x2+10x)3
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
Giair phương trình
a,\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
b,\(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
c,(2x-5)3-(3x-4)3+(x+1)3=0
d,(x2+3x-4)3+(3x2+7x+4)3=(4x2+10x)3
a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)
\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0
\(\Rightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy x = -65
b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)
\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)
Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0
\(\Rightarrow x-99=0\)
\(\Leftrightarrow x=99\)
Vậy x =99
x + 16/49 + x + 18/47 = x + 20/45 - 1
2x +101/141 =x-5/9
<=> 2x-x = -5/9-101/141
<=> x = -538/423