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\(3\sqrt[3]{x-3}+4\sqrt[3]{8x-24}-\frac{1}{3}\sqrt[3]{27x-81}=3\sqrt[3]{x-3}+4\sqrt[3]{8\left(x-3\right)}-\frac{1}{3}\sqrt[3]{27\left(x-3\right)}=3\sqrt[3]{x-3}+4.2.\sqrt[3]{x-3}-\frac{1}{3}.3.\sqrt[3]{x-3}=3\sqrt[3]{x-3}+8\sqrt[3]{x-3}-\sqrt[3]{x-3}=10.\sqrt[3]{x-3}=-20\Leftrightarrow\sqrt[3]{x-3}=-2\Leftrightarrow x-3=-8\Leftrightarrow x=-5\)
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b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)
\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)
Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)
\(\Rightarrow t^2-t-20=0\)
\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)
Thay t=5 vào (1), ta có :
\(\sqrt{x^2-3x+7}=5\)
\(\Leftrightarrow x^2-3x+7=25\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Rightarrow x_1=6;x_2=-3\)
vậy...
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ta có : (x^2+x+4)(1+8x) +16x^2=0
vì 16x^2>=0 suy ra *x^2+x+4=0
*1+8x=0
*16x^2=0
tự giải pt
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2( x - 1 ) - 5 = 3( 5 - 3x)
2x - 2 - 5 = 15 - 9x
2x - 7 = 15 - 9x
2x + 9x = 15 + 7
11x = 22
x = 2
Vậy x = 2
\(2\left(x-1\right)-5=3\left(5-3x\right)\)
\(\Leftrightarrow2x-2-5=15-9x\)
\(\Leftrightarrow2x-\left(2+5\right)=15-9x\)
\(\Leftrightarrow2x-7=15-9x\)
\(\Leftrightarrow2x+9x=15+7\)
\(\Leftrightarrow11x=22\)
\(\Leftrightarrow x=22\div11\)
\(\Leftrightarrow x=2\)
\(\text{Vậy }x=2\)
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