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ĐKXĐ:....
\(\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}\)
\(\Rightarrow4-\sqrt{1-x}=2-x\)
\(\Rightarrow\sqrt{1-x}=2+x\)
\(\Rightarrow1-x=4+4x+x^2\)
\(\Rightarrow1-x-4-4-x^2=0\)
\(\Rightarrow x^2+x+7=0\)
Đến đây dễ rồi làm nốt nha bạn !
ĐKXĐ:....
\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}4−1−x=2−x
\Rightarrow4-\sqrt{1-x}=2-x⇒4−1−x=2−x
\Rightarrow\sqrt{1-x}=2+x⇒1−x=2+x
\Rightarrow1-x=4+4x+x^2⇒1−x=4+4x+x2
\Rightarrow1-x-4-4-x^2=0⇒1−x−4−4−x2=0
\Rightarrow x^2+x+7=0⇒x2+x+7=0
Đến đây dễ rồi làm nốt nha bạn !
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
a) ĐK:\(x\ge4\)
\(\sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}\Leftrightarrow x-1+x-4+2\sqrt{\left(x-1\right)\left(x-4\right)}=x+4\Leftrightarrow9-x=2\sqrt{x^2-5x+4}\left(ĐK:x\le9\right)\Leftrightarrow\left(9-x\right)^2=4\left(x^2-5x+4\right)\Leftrightarrow81-18x+x^2=4x^2-20x+16\Leftrightarrow3x^2-2x-65=0\Leftrightarrow3x^2-15x+13x-65=0\Leftrightarrow3x\left(x-5\right)+13\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(3x+13\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\3x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\left(tm\right)\\x=-\dfrac{13}{3}\left(ktm\right)\end{matrix}\right.\)
Vậy S={5}
b)\(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0\Leftrightarrow\dfrac{2x-1-1}{\left(\sqrt[3]{2x-1}\right)^2+2.\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\left(\sqrt[3]{x-1}\right)^2}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}\right]=0\)(1)
Dễ thấy \(\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}>0\)
Vậy (1)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy S={1}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+8}=x^2+5x+4\left(2\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
(2)\(\Leftrightarrow5\sqrt{a+4}=a\Leftrightarrow25\left(a+4\right)=a^2\Leftrightarrow a^2-25a-100=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=\dfrac{25+5\sqrt{41}}{2}\left(tm\right)\\a=\dfrac{25-5\sqrt{41}}{2}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{25+5\sqrt{41}}{2}\Leftrightarrow\dfrac{25+5\sqrt{41}}{2}=x^2+5x+4\Leftrightarrow25+5\sqrt{41}=2x^2+10x+8\Leftrightarrow2x^2+10x-17-5\sqrt{41}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=3,045972466\left(tm\right)\\x=-8,045972466\left(tm\right)\end{matrix}\right.\)
Vậy S={-8,045972466;3,045972466}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+28}=x^2+5x+4\left(1\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
Vậy \(\left(1\right)\Leftrightarrow5\sqrt{a+24}=a\Leftrightarrow25\left(a+24\right)=a^2\Leftrightarrow a^2-25a-600=0\Leftrightarrow a^2-40a+15a-600=0\Leftrightarrow a\left(a-40\right)+15\left(a-40\right)=0\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-40=0\\a+15=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=40\left(tm\right)\\a=-15\left(ktm\right)\end{matrix}\right.\)
Vậy ta có a=40\(\Leftrightarrow x^2+5x+4=40\Leftrightarrow x^2+5x-36=0\Leftrightarrow x^2-4x+9x-36=0\Leftrightarrow x\left(x-4\right)+9\left(x-4\right)=0\Leftrightarrow\left(x-4\right)\left(x+9\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-4=0\\x+9=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)
Vậy S={-9;4}