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Khách
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KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
25 tháng 8 2020
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
NC
25 tháng 8 2020
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
12 tháng 7 2021
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
17 tháng 9 2023
a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);
b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);
c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) = - 3{x^2}.6{x^2} - - 3{x^2}.8x + - 3{x^2}.1\\ = - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} = - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);
d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);
e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ = - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} = - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);
g)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
C
28 tháng 6 2017
c) \(\left(3x-1\right).\left(2x+7\right)-\left(x+1\right).\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=x+2-x+5\)
\(\Leftrightarrow18x-2-7=0\)
\(\Rightarrow x=\dfrac{9}{18}=\dfrac{1}{2}\)
C
28 tháng 6 2017
b) \(2.\left(3x-1\right).\left(2x+5\right)-6.\left(2x-1\right).\left(x+2\right)=1\)
\(\Leftrightarrow\left(6x-2\right).\left(2x+5\right)-\left(12x-6\right).\left(x+2\right)=1\)
\(\Leftrightarrow12x^2+30x-4x-10-\left(12x^2+24x-6x-12\right)=1\)
\(\Leftrightarrow12x^2+26x-10-12x^2-18x +12=1\)
\(\Leftrightarrow8x+2=1\)
\(\Rightarrow x=\dfrac{-1}{8}\)
28 tháng 5 2022
1: Trường hợp 1: x<-2
Pt sẽ là -x-2+5-x=7
=>-2x+3=7
=>-2x=4
hay x=-2(loại)
Trường hợp 2: -2<=x<5
Pt sẽlà x+2+5-x=7
=>7=7(luôn đúng)
Trường hợp 3: x>=5
Pt sẽ là x+2+x-5=7
=>2x-3=7
=>x=5(nhận)
4: \(\left|x^2-2x\right|=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(x^2-2x\right)^2=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-2x-x\right)\left(x^2-2x+x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-3x\right)\left(x^2-x\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;1;3\right\}\)
5: Ta có: \(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(2x+3+x+2\right)\left(2x+3-x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+5\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};-1\right\}\)
6: |5x-4|=|x+2|
=>5x-4=x+2 hoặc 5x-4=-x-2
=>4x=6 hoặc 6x=2
=>x=3/2 hoặc x=1/3
\(-\left|\dfrac{1}{2}x-2\right|+\dfrac{-2}{3}=\dfrac{-6}{5}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=\dfrac{-6}{5}+\dfrac{2}{3}\\ \Rightarrow-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ \Rightarrow\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=\dfrac{8}{15}\\\dfrac{1}{2}x-2=-\dfrac{8}{15}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{38}{15}\\\dfrac{1}{2}x=\dfrac{22}{15}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{76}{15}\\x=\dfrac{44}{15}\end{matrix}\right.\)
#$\mathtt{Toru}$
\(-\left|\dfrac{1}{2}x-2\right|+\left(-\dfrac{2}{3}\right)=-\dfrac{6}{5}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{6}{5}+\dfrac{2}{3}\\ =>-\left|\dfrac{1}{2}x-2\right|=-\dfrac{8}{15}\\ =>\left|\dfrac{1}{2}x-2\right|=\dfrac{8}{15}\)
TH1: \(\dfrac{1}{2}x-2=\dfrac{8}{15}\left(x\ge4\right)\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{15}+2=\dfrac{38}{15}\\ \Rightarrow x=\dfrac{38}{15}\cdot2=\dfrac{76}{15}\left(tm\right)\)
TH2: \(\dfrac{1}{2}x-2=-\dfrac{8}{15}\left(x< 4\right)\)
\(\Rightarrow\dfrac{1}{2}x=-\dfrac{8}{15}+2=\dfrac{22}{15}\\ \Rightarrow x=\dfrac{22}{15}\cdot2=\dfrac{44}{15}\left(tm\right)\)