\(4\sqrt[3]{x^2-6x+10}-3=1\Leftrightarrow4\sqrt[3]{x^2-6x+10}=4\)

\(\Leftrightarrow\sqrt[3]{x^2-6x+10}=1\Leftrightarrow x^2-6x+10=1\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)

\(4\sqrt[3]{x^2-6x+10}-3=1\)

\(\Leftrightarrow4\sqrt[3]{x^2-6x+10}=4\)

\(\Leftrightarrow\sqrt[3]{x^2-6x+10}=1\)

\(\Leftrightarrow x^2-6x+10=1\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)

a,đk x >= 0  \(\sqrt{16x}=8\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

b, đk x =< 4/5 \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=-140\Leftrightarrow x=-28\)

c;d;e tương tự câu f bạn nhé 

f, đk x >= -1 

\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x=15\)

Câu 33 : \(\sqrt[3]{x^3+3x^2+3x+1}-\sqrt[3]{8x^3+12x^2+6x+1}\)

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

-> chọn B 

Câu 34 : \(\sqrt[3]{x^3-3x^2+3x-1}-\sqrt[3]{125x^3+75x^2+15x+1}\)

\(=\sqrt[3]{\left(x-1\right)^3}-\sqrt[3]{\left(5x+1\right)^3}=x-1-5x-1=-4x-2\)

ta có : \(\hept{\begin{cases}x^3+3x^2+3x+1=\left(x+1\right)^3\\8x^3+12x^2+6x+1=\left(2x+1\right)^3\end{cases}}\)

nên : \(\sqrt[3]{x^3+3x^2+3x+1}-\sqrt[3]{8x^3+12x^2+6x+1}=x+1-\left(2x+1\right)=-x\)

Vậy đáp án là B

a) \(x^2-3x+1=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+3}{2}\\x=\frac{-\sqrt{5}+3}{2}\end{cases}}\)

Vậy \(S=\left\{\frac{\sqrt{5}+3}{2};\frac{-\sqrt{5}+3}{2}\right\}\)

b) \(x^2+2x+1=1\)

\(\Leftrightarrow x^2+2x=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Vậy \(S=\left\{0;-2\right\}\)

x²-3x+1=0

x²-3x+2,25=1,25

(x-1,5)²=1,25

x-1,5=\(\pm\sqrt{1,25}\)

x=\(\pm\sqrt{1,25}\)+1,5

x²+2x+1=1

x(x+2)=0

x=0 hoặc x+2=0 suy ra x=-2

vậy x=0 và -2

\(\Delta=b^2-4ac=3^2-4=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

x²+3x+1=0

x²+3x+2,25=1,25

(x+1,5)²=1,25

x+1,5=\(\pm\sqrt{1,25}\)

x=\(\pm\sqrt{1,25}\)-1,5

a, ĐK :a >= 3

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)

b, \(ĐK:x\ge-\frac{1}{2}\)

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\sqrt{2x+1}=3\)

\(\Leftrightarrow x=4\left(tm\right)\)

a) đk: \(a\ge3\)

pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)

\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)

\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)

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

Đặt \(a=2x-3;b=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\)

Ta có hpt \(\hept{\begin{cases}a^2+x-3=\left(x-1\right)b\\b^2+x-3=\left(x-1\right)a\end{cases}}\)

Trừ 2 pt trên ta được: \(a^2-b^2=\left(x-1\right)\left(b-a\right)\Rightarrow\left(a-b\right)\left(a+b+x-1\right)=0\)

+) Nếu \(a=b\Leftrightarrow2x-3=\sqrt{2x^2-6x+6}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\2x^2-6x+3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3-\sqrt{3}}{2}\left(loại\right)\\x=\frac{3+\sqrt{3}}{2}\left(tm\right)\end{cases}}}\)

+) Nếu \(2x-3+\sqrt{2x^2-6x+6}+x-3=0\Leftrightarrow\sqrt{2x^2-6x}=6-3x\)\(\Leftrightarrow\hept{\begin{cases}x\le2\\7x^2-30x+36=0\end{cases}\left(VN\right)}\)

Vậy pt có nghiệm duy nhất: \(x=\frac{3+\sqrt{3}}{2}\)

\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)

\(\Leftrightarrow\left(4x^2-12x+9\right)+x-3=\left(x-1\right)\sqrt{2x^2-5x+3-x+3}\)

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

đặt \(\hept{\begin{cases}t=2x-3\\y=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\left(y\ge0\right)\end{cases}}\)

ta có hệ : \(\hept{\begin{cases}t^2+x-3=\left(x-1\right)y\\y^2-\left(x-1\right)t+x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}t^2-\left(x-1\right)y+\left(x-3\right)=0\\y^2-\left(x-1\right)t+\left(x-3\right)=0\end{cases}}\)

\(\Rightarrow t^2-y^2-\left(x-1\right)y+\left(x-1\right)t=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y\right)+\left(x-1\right)\left(t-y\right)=0\)

\(\Leftrightarrow\left(t-y\right)\left(t+y+x-1\right)=0\)

th1 : \(t-y=0\Leftrightarrow t=y\Leftrightarrow2x-3=\sqrt{\left(x-1\right)\left(2x-3\right)-\left(x-3\right)}\) ĐK : \(x\ge\frac{3}{2}\)

\(\Leftrightarrow4x^2-12x+9=2x^2-6x+6\)

\(\Leftrightarrow2x^2-6x+3=0\)

\(\Delta=b^2-4ac=\left(-6\right)^2-4\cdot2\cdot3=12\)

\(\Rightarrow\orbr{\begin{cases}x=3+\sqrt{3}\left(tm\right)\\x=3-\sqrt{3}\left(loai\right)\end{cases}}\)

th2 : \(x+y+t-1=0\Leftrightarrow y=1-x-t\)

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

\(\Leftrightarrow\sqrt{2x^2-6x+6}=4-3x\left(đk:x\le\frac{4}{3}\right)\)

\(\Leftrightarrow2x^2-6x+6=16-24x+9x^2\)

\(\Leftrightarrow7x^2-18x+10=0\)

\(\Delta=b^2-4ac=\left(-18\right)^2-4\cdot7\cdot10=44\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{18+\sqrt{44}}{2}=9+\sqrt{11}\left(loai\right)\\x=\frac{18-\sqrt{44}}{2}=9-\sqrt{11}\left(loai\right)\end{cases}}\)

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