Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Hoàng Như Quỳnh]

\(a,\sqrt{\left(2x-3\right)^2}=4\)

\(\left|2x-3\right|=4\)

\(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}\)

\(b,\sqrt{4x^2-12x+13}=2\)

\(\sqrt{4x^2-12x+13}^2=2^2\)

\(4x^2-12+13>0\)

\(\left|4x^2-12x+13\right|=4\)

\(4x^2-12x+13=4\)

\(4x^2-12x+9=0\)

\(\left(2x-3\right)^2=0< =>2x-3=0< =>x=\frac{3}{2}\)

\(c,\sqrt{3x^2-9x+1}=x-2\)

\(3x^2-9x+1=x^2-4x+4\)

\(2x^2-5x-3=0\)

\(2x^2-6x+1x-3=0\)

\(2x\left(x-3\right)+\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+1\right)=0\)

\(\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\)

Answer 2

[Hoàng Như Quỳnh]

\(d,3x^2+4x+1=961-124x+4x^2\)

\(128x-x^2=960\)

\(x^2-128x+960=0\)

\(\sqrt{\Delta}=\sqrt{\left(-128\right)^2-\left(4.1.960\right)}=112\)

\(x_1=\frac{128+112}{2}=120\left(tm\right)\)

\(x_2=\frac{128-112}{2}=8\left(TM\right)\)

\(e,\sqrt{x+4\sqrt{x-4}}=2\)

\(\sqrt{x-4+4\sqrt{x-4}+4}=2\)

\(\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)

\(\left|\sqrt{x-4}+2\right|=2\)

\(\orbr{\begin{cases}\sqrt{x-4}+2=2\\\sqrt{x-4}+2=-2\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\\sqrt{x-4}=-4\left(KTM\right)\end{cases}}}\)

\(f,\sqrt{x-1+6\sqrt{x-1}+9}=5\)

\(\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)

\(\left|\sqrt{x-1}+3\right|=5\)

\(\orbr{\begin{cases}\sqrt{x-1}+3=5\\\sqrt{x-1}+3=-5\end{cases}\orbr{\begin{cases}x=5\\\sqrt{x-1}=-8\left(KTM\right)\end{cases}}}\)

\(g,\sqrt{4x^2+4x+1}=\sqrt{x^2-6x+9}\)

\(\sqrt{\left(2x+1\right)^2}=\sqrt{\left(x-3\right)^2}\)

\(\left|2x+1\right|=\left|x-3\right|\)

\(\orbr{\begin{cases}2x+1=x-3\\2x+1=3-x\end{cases}\orbr{\begin{cases}x=-4\\x=\frac{2}{3}\end{cases}}}\)

\(h,\sqrt{\left(x-3\right)^2}+\sqrt{\left(x-5\right)^2}=10\)

\(\left|x-3\right|+\left|x-5\right|=10\)

lập bảng xét dấu rồi chia TH

\(TH1:x\le3\)

\(3-x+5-x=10\)

\(x=-1\left(TM\right)\)

\(TH2:3< x\le5\)

\(x-3+5-x=10\)

\(0x=8\left(KTM\right)\)

\(TH3:x>5\)

\(x-3+x-5=10\)

\(x=9\left(TM\right)\)

HOK tốt