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a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a) \(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{4}-\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{15}{20}-\frac{12}{20}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{13}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{10}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\)
Vậy: \(x=\frac{10}{3}\)
b) \(3x.\left(\frac{1}{2}.x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
Vậy: \(x\in\left\{0;2\right\}\)
c) \(\left(4-x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{3}{2}\end{cases}}}\)
Vậy: \(x\in\left\{4;\frac{3}{2}\right\}\)
d) \(\frac{4}{-3}=\frac{-12}{x}\)
\(\Leftrightarrow4x=\left(-12\right).\left(-3\right)\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
Vậy: \(x=9\)
e) \(\frac{4x}{-3}=\frac{12}{-x}\)
\(\Leftrightarrow4x.\left(-x\right)=12.\left(-3\right)\)
\(\Leftrightarrow-4x^2=-36\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy: \(x\in\left\{3;-3\right\}\)
1/
a, \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b, \(B=2\dfrac{3}{11}.\dfrac{11}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{11}{12}.\left(-\dfrac{11}{5}\right)=-\dfrac{55}{12}\)
c, \(C=\left(\dfrac{3}{4}-0,2\right):\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right):\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}:\left(-\dfrac{2}{5}\right)=-\dfrac{11}{8}\)
2/
a, \(\dfrac{11}{12}-x=\dfrac{2}{3}+\dfrac{1}{4}\\ \Rightarrow\dfrac{11}{12}-x=\dfrac{11}{12}\\ \Rightarrow x=0\)
b, \(2x\left(x-\dfrac{1}{7}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\\ \Rightarrow x=-\dfrac{5}{7}\)
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
a, Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{-2}=\frac{2x+5y}{2.3+5.\left(-2\right)}=-\frac{12}{-4}=3\)
\(x=-3;y=6\)
b, Theo bài ra ta có : \(x:y=4:5\Leftrightarrow\frac{x}{4}=\frac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}=\frac{y}{5}=\frac{x-y}{4-5}=\frac{13}{-1}=-13\)
\(x=-52;y=-65\)
c, Theo bài ra ta có: \(4x=7y\Leftrightarrow\frac{x}{7}=\frac{y}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{7}=\frac{y}{4}=\frac{x-y}{7-4}=\frac{12}{3}=4\)
\(x=28;y=16\)
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
Giải :
\(\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow4.\left(x-1\right)=3.\left(x-2\right)\)
\(\Rightarrow4x-4=3x-6\)
\(\Rightarrow4x-4-3x+6=0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)Không thỏa mãn => Không có giá trị x thỏa mãn đề bài
\(\frac{2x-3}{x+1}=\frac{4}{7}\)
\(\Rightarrow7.\left(2x-3\right)=4.\left(x+1\right)\)
\(\Rightarrow14x-21-4x-4=0\)
\(\Rightarrow10x-25=0\)
\(\Rightarrow10x=25\)
\(\Rightarrow x=\frac{25}{10}=\frac{5}{2}\)
Giá trị trên thỏa mãn đầu bài
Các phần khác em làm tương tự nha
Trả lời :
a, \(\frac{3}{4}-\left(\frac{1}{2}\div x+\frac{1}{2}\right)=\frac{3}{5}\)
=> \(\frac{1}{2}\div x+\frac{1}{2}=\frac{3}{20}\)
=> \(\frac{1}{2}\div x=\frac{-7}{20}\)
=> \(x=\frac{-10}{7}\)
b, (4 - x) . (2x + 3) = 0
=> \(\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=\frac{-3}{2}\end{cases}}\)
c, \(\frac{4}{-3}=\frac{-12}{x}\)
=> 4x = 36
=> x = 9
d, \(\frac{4x}{-3}=\frac{12}{-x}\)
=> \(-4x^2=-36\)
=> 4x2 = 36
=> x2 = 9
=> x = \(\pm3\)