a. |5x+4|+7=26
b
b. 3-4|5-6x|=7
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\(a,\left|5x+4\right|+7=26\\ \left|5x+4\right|=26+7\\ \left|5x+4\right|=33\\ \Rightarrow\left\{{}\begin{matrix}5x+4=33\\5x+4=-33\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x=29\\5x=-29\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\\x=-\dfrac{29}{5}\end{matrix}\right.\)
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Answer:
\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)
\(\Rightarrow6x^2-6x^2-5x+6=7\)
\(\Rightarrow-5x+6=7\)
\(\Rightarrow-5x=1\)
\(\Rightarrow x=\frac{-1}{5}\)
\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)
\(\Rightarrow5x.19-240x^2+15x=-100\)
\(\Rightarrow95x-240x^2+15x=-100\)
\(\Rightarrow-240x^2+110x+100=0\)
\(\Rightarrow-24x^2-11x-10=0\)
\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)
\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)
\(\left(3x-5\right)\left(7-5x\right)-\left(5x-2\right)\left(2-3x\right)=4\)
\(\Rightarrow\left(21x-15x^2-35+25x\right)-\left(10x-15x^2-4+6x\right)-4=0\)
\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)
\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)
\(\Rightarrow30x-35=0\)
\(\Rightarrow x=\frac{7}{6}\)
a: =3x^3-15x^2+21x
b: =-x^3+6x^2+5x-4x^2-24x-20
=-x^3+2x^2-19x-20
c: =9x^2+15x-3x-5-7x^2-14
=2x^2+12x-19
d: =10x^2-4x+2/3
a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)
=>-5x=1
hay x=-1/5
b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)
\(\Leftrightarrow-180x^2+50x+100=0\)
hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)
c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)
=>30x-31=4
=>30x=35
hay x=7/6
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)
=> -5(6x - 5) = -7(5x - 3)
=> -30x + 25 = -35x + 21
=> -30x + 25 + 35x - 21 = 0
=> (-30x + 35x) + (25 - 21) = 0
=> 5x + 4 = 0
=> 5x = -4
=> x = -4/5
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
=> -5(12 - 7x) = -13(4 - 3x)
=> -60 + 35x = -52 + 39x
=> -60 + 35x + 52 - 39x = 0
=> (-60 + 52) + (35x - 39x) = 0
=> -8 - 4x = 0
=> -8 = 4x
=> x = -2
c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)
=> 15(2x + 4) = 7(4x - 2)
=> 30x + 60 = 28x - 14
=> 30x + 60 - 28x + 14 = 0
=> 2x + 74 = 0
=> 2x = -74
=> x = -37
a) 71 + \(\frac{26-3x}{5}=75\)
↔ \(\frac{26-3x}{5}=4\)
↔\(26-3x=20\)
↔x=2
b) \(x-17=\left(-8\right)-17\)
↔ \(x-17=-25\)
↔\(x=-8\)
c)\(\left(5x-2\right).6^3=3.6^5\)
↔\(\left(5x-2\right)=3.6^2\)
↔\(5x=110\)
↔\(x=22\)
d)\(\left(3x-2^4\right).7^3=2.7^4\)
↔\(3x-16=14\)
↔x=10
tick nha~~
a. |5x+4|+7=26
=>|5x+4|=19
=>5x+4=19 hoặc -19
=>5x=15
=>x=3
=>5x=-23
=>x=-23/5
b. 3-4|5-6x|=7
=>4|5-6x|=-4
=>|5-6x|=-1 (vô nghiệm)
Giá trị tuyệt đối của 1 số luôn >=0
a) |5x+4| +7 = 26
|5x+4| = 19
Th1:
5x +4 = 19
5x= 15
x=3
Th2:
5x +4 = -19
5x = -23
x= -23/5
Vậy x= 3 và x=-23/5
b) 3-4|5-6x|=7
4|5-6x|= -4
|5-6x|= -1
=> x thuộc rỗng.