M.n júp mk vs ạ.
Tìm x biết
3.(x+5)-4.(x+7)=20.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)
<=> 13(x + 1) - 2(5x + 3) = x + 7
<=> 13x + 13 - 10x - 6 = x + 7
<=> 3x + 7 = x + 7
<=> 3x + 7 - x = 7
<=> 2x + 7 = 7
<=> 2x = 7 - 7
<=> 2x = 0
<=> x = 0
c) 2x + 4(x - 2) = 5
<=> 2x + 4x - 8 = 5
<=> 6x - 8 = 5
<=> 6x = 5 + 8
<=> 6x = 13
<=> x = 13/6
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)
\(\Rightarrow\left[x^2+4x-5\right]\left[x^2+4x-5-16\right]=297\)
Đặt \(x^2+4x-5=t\)
\(\Rightarrow t\left(t-16\right)=297\)
\(\Rightarrow t^2-16t+64=297+64\)
\(\Rightarrow\left(t+8\right)^2=361\)
\(\Rightarrow\left[{}\begin{matrix}t+8=19\\t+8=-19\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t=11\\t=-27\end{matrix}\right.\)
Ta có : \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)
\(\Leftrightarrow\left(x^2-x+5x-5\right)\left(x^2-3x+7x-21\right)=297\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)
\(\Leftrightarrow\left(x^2+4x-13+8\right)\left(x^2+4x-13-8\right)=297\)
\(\Leftrightarrow\left(x^2+4x-13\right)^2-64=297\)
\(\Leftrightarrow\left(x^2+4x-13\right)^2=361\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x-13=19\\x^2+4x-13=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+4-17=19\\x^2+4x+4-17=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2-17=19\\\left(x+2\right)^2-17=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2=36\\\left(x+2\right)^2=-2\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
\(3\left(x-2\right)+4\left(x-5\right)=23\)
\(\Rightarrow3x-6+4x-20-23=0\)
\(\Rightarrow7x-49=0\)
\(\Rightarrow x=7\)
3(x-2)+4(x-5)=23
<=>3x-6+4x-20=23
<=>7x-26=23
<=>7x=49
<=>x=7
Vậy x=7
\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)
\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}=\frac{20-9}{36}=\frac{11}{36}\)
\(x=\frac{3}{7}:\frac{11}{36}=\frac{3}{7}\times\frac{36}{11}=\frac{3\times36}{7\times11}=\frac{108}{77}\)
Vậy \(x=\frac{108}{77}\)
\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)
\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}\)
\(\frac{3}{7}:x=\frac{11}{36}\)
\(x=\frac{3}{7}:\frac{11}{36}\)
\(x=\frac{108}{77}\)
a) \(\frac{35}{14}=\frac{5\times7}{2\times7}=\frac{5}{2}\)
\(\frac{125}{50}=\frac{5\times5\times5}{2\times5\times5}=\frac{5}{2}\)
b)\(\frac{4\times5+4\times11}{8\times7+4\times3}=\frac{4\times\left(5+11\right)}{4\times\left(2\times7+3\right)}=\frac{16}{17}\)
c) \(\frac{3\times11+7\times11}{22\times2+11\times6}=\frac{11\times\left(3+7\right)}{22\times\left(2+3\right)}=\frac{11\times10}{22\times5}=\frac{11\times2\times5}{11\times2\times5}=1\)
(x-1/2)=19/4:5/3
(x-1/2)=19/4x3/5
x-1/2=57/20
x=57/20+1/2
x=67/20
tk cho mk nha
\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)
\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)
\(x=\frac{57}{20}+\frac{1}{2}\)
\(x=\frac{67}{20}\)
3. (x+5) - 4.(x+7)=20
3x +15-4x-28 . =20
-x . =20+28-15
-x . =33
x. -33
3(x+5)-4(x+7)=20
<=> 3x+15-4x-28=20
<=> -x-13=20
<=> x+13=-20
<=> x=-33