|x−1|−2x=5
tìm x ạ !! nhanh giùm
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\(C=\left(x+1\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi x=-1
\(\left(2x-1\right)^3=27\)
\(\left(2x-1\right)^3=3^3\)
\(2x-1=3\)
\(2x=3+1=4\)
\(x=4:2\)
\(x=2\)
\(ĐKXĐ:x\ne\pm5\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow\frac{3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}+\frac{30}{4\left(25-x^2\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15}{4\left(x-5\right)\left(x+5\right)}+\frac{-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x-15}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3}{4\left(x+5\right)}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow18\left(x+5\right)=-28\left(x+5\right)\)
\(\Rightarrow18\left(x+5\right)+28\left(x+5\right)=0\)
\(\Rightarrow46\left(x+5\right)=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\)(ktm)
Vậy pt vô nghiệm
\(\left(4x-5\right)\left(2x+30\right)-4\left(x+2\right)\left(2x-1\right)+\left(10x+7\right)\)
\(=8x^2+110x-150-8x^2-12x+8+10x+7\)
\(=108x-135\)
2x+1 \(⋮\)x-2
=> (2x-4)+5 \(⋮\)x-2
=> 2(x-2) + 5\(⋮\)x-2
=> 5 \(⋮\)x-2 ( 2(x-2) \(⋮\)x-2)
=> x-2 \(\inƯ\left(5\right)\)
=> x-2 \(\in\){1;5} ( x>2 => x-2 >0)
=> x \(\in\){3;7}
Trường hợp 1: x - 1 ≥ 0 → x ≥ 1
→ x - 1 = 2x - 5
→ x - 2x = -5 + 1
→ - x = - 4
→ x = 4
Trường hợp 2: x - 1 ≤ 1 → x ≤ 1
→ - ( x - 1) = 2x - 5
→ - x + 1 = 2x - 5
→ -x - 2x = -5 - 1
→ -3x = 6
→ x = 2 (loại)
Vậy, x = 4