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a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
a: \(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
=>14x=7
hay x=1/2
b: \(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
=>-56x+156=24x-324
=>-80x=-480
hay x=6
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
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: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
\(\sqrt{\left(3x-1\right)^2}=\sqrt{\left(15-2x\right)^2}\) " phá trị tuyệt đối = bình phương trong căn
\(\left(3x-1\right)^2=\left(15-2x\right)^2\) " phá căn = bình phương 2 vế
\(\left(9x^2-6x+1\right)=225-60x+4x^2\)
\(5x^2+54x-224=0\)
\(\Delta=\frac{54^2-4.\left(-224.5\right)}{5}=7396\)
\(\hept{\begin{cases}x_1=\frac{-54+\sqrt{7396}}{10}=\frac{-54+86}{10}=\frac{32}{10}=\frac{16}{5}\\x_2=\frac{-54-86}{10}=\frac{-140}{10}=-14\end{cases}}\)
\(15-2\left(2x+1\right)=8+3x\\ \Leftrightarrow15-4x-2=8+3x\\ \Leftrightarrow-4x-3x=8-15+2\\ \Leftrightarrow-7x=-5\\ x=\dfrac{5}{7}\)
\(a,\Rightarrow x\in\varnothing\left(\left|4+2x\right|\ge0>-4\right)\\ b,\Rightarrow\left|3x-1\right|=x-2\\ \Rightarrow\left[{}\begin{matrix}3x-1=x-2\left(x\ge\dfrac{1}{3}\right)\\3x-1=2-x\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x\in\varnothing\\ c,\Rightarrow\left|x+15\right|=3x-1\\ \Rightarrow\left[{}\begin{matrix}x+15=3x-1\left(x\ge-15\right)\\x+15=1-3x\left(x< -15\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-\dfrac{7}{2}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x=8\)
|2x-1|+3x=15
Ta có: |2x-1|=2x-1 <=> 2x-1 \(\ge\) 0 <=> x \(\ge\) 1/2
|2x-1|=-(2x-1)=-2x+1 <=> -2x+1 < 0 <=> -2x<-1 <=> x < 1/2
Nếu x \(\ge\) 1/2 thì (1) <=> 2x-1+3x=15 <=> 5x=16 <=> x=16/5
Nếu x < 1/2 thì (1) <=> -2x+1+3x=15 <=> x=14
Vậy x \(\in\) {16/5;14}
sai đoạn giữa rồi hoàng phúc ơi