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Question

Giúp mình phần b bài 2 và bài 3

Trần Tuấn Hoàng · Accepted Answer

Bài 2:$\left|\left|x^3-4\right|+21\right|:5=5$$\Leftrightarrow\left|\left|x^3-4\right|+21\right|=25$$\Leftrightarrow\left|x^3-4\right|+21=25$ hay $\left|x^3-4\right|+21=-25$$\Leftrightarrow\left|x^3-4\right|=4$ hay $\left|x^3-4\right|=-46$ (vô lí do $\left|x^3-4\right|\ge0\forall x$)$\Leftrightarrow x^3-4=4$ hay $x^3-4=-4$$\Leftrightarrow x^3-8=0$ hay $x^3=0$$\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)=0$ hay $x=0$$\Leftrightarrow\left(x-2\right)\left(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right)=0$ hay $x=0$$\Leftrightarrow\left(x-2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0$ hay $x=0$$\Leftrightarrow x=2$ hay $x=0$ hay $\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0$ (vô nghiệm do $\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0$)-Vậy $S=\left\{0;2\right\}$Câu trả lời: Bài 2:$\left|\left|x^3-4\right|+21\right|:5=5$$\Leftrightarrow\left|\left|x^3-4\right|+21\right|=25$$\Leftrightarrow\left|x^3-4\right|+21=25$ hay $\left|x^3-4\right|+21=-25$$\Leftrightarrow\left|x^3-4\right|=4$ hay $\left|x^3-4\right|=-46$ (vô lí do $\left|x^3-4\right|\ge0\forall x$)$\Leftrightarrow x^3-4=4$ hay $x^3-4=-4$$\Leftrightarrow x^3-8=0$ hay $x^3=0$$\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)=0$ hay $x=0$$\Leftrightarrow\left(x-2\right)\left(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right)=0$ hay $x=0$$\Leftrightarrow\left(x-2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0$ hay $x=0$$\Leftrightarrow x=2$ hay $x=0$ hay $\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0$ (vô nghiệm do $\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0$)-Vậy $S=\left\{0;2\right\}$

Trần Tuấn Hoàng · Answer

Bài 3:$\left|\left|2x^2-2\right|+6\left|x^2-1\right|\right|=4^6:\left(2^3\right)^2$$\Leftrightarrow\left|\left|2x^2-2\right|+6\left|x^2-1\right|\right|=64$$\Leftrightarrow\left|2x^2-2\right|+6\left|x^2-1\right|=64$ (*) hay $\Leftrightarrow\left|2x^2-2\right|+6\left|x^2-1\right|=-64$ (pt vô nghiệm do $\left|2x^2-2\right|+6\left|x^2-1\right|$ luôn là số thực dương)-Có: $\left|2x^2-2\right|=2x^2-2$ nếu $x\ge1$ hay $x\le-1$.$\left|2x^2-2\right|=-2x^2+2$ nếu $x\le1$ hay $x\ge-1$.$6\left|x^2-1\right|=6\left(x^2-1\right)$ nếu $x\ge1$ hay $x\le-1$$6\left|x^2-1\right|=-6\left(x^2-1\right)$ nếu $x\le1$ hay $x\ge-1$-TH1: $x\le-1$:(*) $\Leftrightarrow2x^2-2+6\left(x^2-1\right)=64$$\Leftrightarrow2x^2-2+6x^2-6=64$$\Leftrightarrow8x^2-72=0$$\Leftrightarrow x^2-9=0$ $\Leftrightarrow\left(x-3\right)\left(x+3\right)=0$$\Leftrightarrow x=3$ (loại) hay $x=-3$ (nhận)-TH2: $-1\le x\le1$:(*) $\Leftrightarrow-2x^2
+2-6\left(x^2-1\right)=64$$\Leftrightarrow-2x^2+2-6x^2
+6=64$$\Leftrightarrow-8x^2-56=0$$\Leftrightarrow8x^2+56=0$ (pt vô nghiệm do $8x^2+56\ge56\forall x$)-TH3: $x\ge1$:-TH1: $x\le-1$:(*) $\Leftrightarrow2x^2-2+6\left(x^2-1\right)=64$$\Leftrightarrow2x^2-2+6x^2-6=64$$\Leftrightarrow8x^2-72=0$$\Leftrightarrow x^2-9=0$ $\Leftrightarrow\left(x-3\right)\left(x+3\right)=0$$\Leftrightarrow x=3$ (nhận) hay $x=-3$ (loại)-Vậy $S=\left\{3;-3\right\}$Câu trả lời: Bài 3:$\left|\left|2x^2-2\right|+6\left|x^2-1\right|\right|=4^6:\left(2^3\right)^2$$\Leftrightarrow\left|\left|2x^2-2\right|+6\left|x^2-1\right|\right|=64$$\Leftrightarrow\left|2x^2-2\right|+6\left|x^2-1\right|=64$ (*) hay $\Leftrightarrow\left|2x^2-2\right|+6\left|x^2-1\right|=-64$ (pt vô nghiệm do $\left|2x^2-2\right|+6\left|x^2-1\right|$ luôn là số thực dương)-Có: $\left|2x^2-2\right|=2x^2-2$ nếu $x\ge1$ hay $x\le-1$.$\left|2x^2-2\right|=-2x^2+2$ nếu $x\le1$ hay $x\ge-1$.$6\left|x^2-1\right|=6\left(x^2-1\right)$ nếu $x\ge1$ hay $x\le-1$$6\left|x^2-1\right|=-6\left(x^2-1\right)$ nếu $x\le1$ hay $x\ge-1$-TH1: $x\le-1$:(*) $\Leftrightarrow2x^2-2+6\left(x^2-1\right)=64$$\Leftrightarrow2x^2-2+6x^2-6=64$$\Leftrightarrow8x^2-72=0$$\Leftrightarrow x^2-9=0$ $\Leftrightarrow\left(x-3\right)\left(x+3\right)=0$$\Leftrightarrow x=3$ (loại) hay $x=-3$ (nhận)-TH2: $-1\le x\le1$:(*) $\Leftrightarrow-2x^2
+2-6\left(x^2-1\right)=64$$\Leftrightarrow-2x^2+2-6x^2
+6=64$$\Leftrightarrow-8x^2-56=0$$\Leftrightarrow8x^2+56=0$ (pt vô nghiệm do $8x^2+56\ge56\forall x$)-TH3: $x\ge1$:-TH1: $x\le-1$:(*) $\Leftrightarrow2x^2-2+6\left(x^2-1\right)=64$$\Leftrightarrow2x^2-2+6x^2-6=64$$\Leftrightarrow8x^2-72=0$$\Leftrightarrow x^2-9=0$ $\Leftrightarrow\left(x-3\right)\left(x+3\right)=0$$\Leftrightarrow x=3$ (nhận) hay $x=-3$ (loại)-Vậy $S=\left\{3;-3\right\}$

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