tính giá trị của biểu thức sau:
a) \(\dfrac{m^2.\left(m+n^2\right).\left(m^3-n^2\right).\left(m^2-n\right)}{m^2+n^2}\)
b) \(a^3.\left(a+b\right).\left(a^5-b^5\right).\left(a^2-b\right)\)
tại a = 5 , b = 25
Cho 2 đa thức: \(N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\)
\(M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\)
a) Tính \(N\left(x\right)-M\left(x\right)\)
b) Tính \(M\left(x\right)+N\left(x\right)\)
a)
\(\begin{matrix}N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\^-M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\\overline{N\left(x\right)-M\left(x\right)=-3x^4+18x^3-2x^2-4x-1}\end{matrix}\)
b)
\(\begin{matrix}M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\^+N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\\overline{M\left(x\right)+N\left(x\right)=-5x^4+14x+\dfrac{5}{3}}\end{matrix}\)
1 (5 điểm)
a) Tính giá trị biểu thức: \(L=\left(-\dfrac{3}{4}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
b) Tính giá trị nhỏ nhất của biểu thức: \(L=\left[\left(x+1\right)^2+3\right]^2+\left|y-5\right|+2008\)
2(4 điểm)
a) Tìm 3 số x;y;z thỏa mãn \(20x=15y=12z\) và \(2x^2+2y^2-3z^2=-100\)
b) Cho đa thức \(L_1\left(x\right)=x^2+2xm+m^2\) và \(L_2\left(x\right)=x^2+\left(2x+1\right)x+m^2\)
Tìm m biết \(L_1\left(1\right)=L_2\left(-1\right)\)
3(4 điểm)
a) Chứng minh \(5^{n+3}-3^{n+3}+5^{n+2}-3^{n+1}⋮60\) với mọi n thuộc N
b) Chứng minh \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2017}{4^{2017}}< \dfrac{1}{2}\)
6 điểm được free ạ =)))))
Cho a,b,c > 0 thoả mãn a+b+c=1 . Tìm giá trị nhỏ nhất của biểu thức:
\(P=\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{c^3}{\left(1-c\right)^2}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\)
Vậy \(P_{min}=\dfrac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Xét các số dương a,b,c thỏa mãn điều kiện abc = 1. Tìm giá trị lớn nhất của biểu thức: \(P=\dfrac{1}{\left(a^2+1\right)\left(b+c\right)}+\dfrac{1}{\left(b^2+1\right)\left(c+a\right)}+\dfrac{1}{\left(c^2+1\right)\left(a+b\right)}\)
Áp dụng BĐT AM-GM: \(VT\le\sum\dfrac{1}{\sqrt{a^2+1}.\sqrt{2a}.2\sqrt{bc}}=\sum\dfrac{1}{2\sqrt{2}\sqrt{a^2+1}}\)
Ta đi chứng minh \(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
Giả sử c=max{a, b, c}.Suy ra \(c\ge1\) nên \(ab\le1\). Ta có bổ đề:
\(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}\le\dfrac{2}{\sqrt{1+ab}}\)(*)
#cm: Áp dụng Bunyakovsky: \(VT_{(*)} \)\(\le\sqrt{2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\right)}\)
Xét \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}-\dfrac{2}{ab+1}=\dfrac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\le0\)
Nên \(VT_{(*)}\)\(\le\sqrt{2.\dfrac{2}{ab+1}}=\dfrac{2}{\sqrt{ab+1}}\), suy ra đpcm.
Do đó \(VT\le\dfrac{2}{\sqrt{ab+1}}+\dfrac{1}{\sqrt{c^2+1}}=2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\)
# cm: \(2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{2c\left(c^2+1\right)}+\sqrt{2c+2}\le3\sqrt{\left(c+1\right)\left(c^2+1\right)}\)
\(\Leftrightarrow8c^3+10c+2+8\sqrt{c\left(c+1\right)\left(c^2+1\right)}\le9\left(c^3+c^2+c+1\right)\)
hay \(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le c^3+9c^2-c+7\) ($)
Áp dụng BĐT AM-GM cho VT của ($):
\(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le4\left(2c^2+c+1\right)\) .Ta chứng minh
\(8c^2+4c+4\le c^3+9c^2-c+7\) hay \(\left(c-1\right)^2\left(c+3\right)\ge0\) (đúng)
Vậy ta có đpcm. Dấu = xảy ra khi a=b=c=1
giúp mình với nhanh nha, mai nộp rồi!!!
1. Tính giá trị của biểu thức:
\(A=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right)\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)\)
biết \(m+n+p=0\)
2. Tính:
a) \(A=\frac{2^3+1}{2^3-1}.\frac{3^3+1}{3^3-1}.\frac{4^3+1}{4^3-1}...\frac{10^3+1}{10^3-1}\)
b) \(B=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(9^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(10^4+\frac{1}{4}\right)}\)
bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)
Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)
\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)
\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)
\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)
\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)
\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)
Vì m+n+p=0=>m+n=-p
\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)
\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)
\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)
\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)
\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)
\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)
\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)
\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)
\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)
\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)
\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)
Từ (1),(2),(3) suy ra :
\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)
\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)
*Tới đây để tính được m3+n3+p3,ta cần CM được bài toán phụ sau:
Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)
Từ m+n+p=0=>m+n=-p
Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)
\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)
Vậy ta đã CM được bài toán phụ
*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)
Vậy A=9
bài 2)
a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:
\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)
\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)
suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)
Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)
\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)
...........................
\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)
\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)
\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)
Vậy A=2036/37
b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 14=1 mà
Nhận thấy các thừa số của B có dạng tổng quát:
\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)
\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)
Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)
Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)
Vậy B=1/221
Bài 1:
Cho phân thức: \(M=\dfrac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+bc+ca\right)^2}{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}\)
a, tìm các giá trị của a, b, c để phân thức được xác định (tức là để mẫu khác 0)
b, Rút gọn M
Bài 2: Rút gọn:
\(A=\dfrac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)
Bài 3: CMR: với mọi số nguyên n thì phân số \(\dfrac{n^3+2n}{n^4+3n^2+1}\) là phân số tối giản
Bài 4: CMR: \(1+x+x^2+x^3+...+x^{31}=\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
Mn giúp mik vs ạ :((
Bài 1:
a, Ta có:
\(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\Leftrightarrow a+b=b+c=c+a=0\)
\(\Leftrightarrow a=b=c=0\)
Vậy điều kiện để phân thức M được xác định là a, b, c không đồng thời = 0
b, Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
Đặt: \(a^2+b^2+c^2=x,ab+bc+ca=y\)
=> \(\left(a+b+c\right)^2=x+2y\)
Ta cũng có:
\(M=\dfrac{x\left(x+2y\right)+y^2}{x+2y-y}=\dfrac{x^2+2xy+y^2}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
\(=a^2+b^2+c^2+ab+bc+ca\)
cho 3 số a,b,c thỏa mãn
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
tính giá trị biểu thức A=\(\dfrac{a^2+b^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{b^2+c^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{c^2+a^2}{\left(c+b\right)\left(a+b\right)}\)
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
\(\Leftrightarrow\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=a+b+c\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{a+b+c}{abc}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)
\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Thay vào A r tính thôi
tìm số tự nhiên m thỏa mãn đồng thời cả 2 ptrình sau:
a, \(4\left(n+1\right)+3n-6< 19\)
b, \(\left(n-3\right)^2-\left(n+4\right)\left(n-4\right)\le43\)
Với giá trị nào của m thì biểu thức:
a,\(\dfrac{m-2}{4}+\dfrac{3m+1}{3}\)giá trị âm
b, \(\dfrac{m-4}{6m+9}\)có giá trị dương
c,\(\dfrac{2m-3}{2m+3}+\dfrac{2m+3}{2m-3}\) có giá trị âm
d, \(\dfrac{-m+1}{m+8}+\dfrac{m-1}{m+3}\) có giá trị dương
e,\(\dfrac{\left(m+1\right)\left(m-5\right)}{2}\)
bài 1:
a) 4n+4+3n-6<19
<=> 7n-2<19
<=> 7n<21 <=> n< 3
b) n\(^2\) - 6n + 9 - n\(^2\) + 16\(\leq\)43
-6n+25\(\leq\)43
-6n\(\leq\)18
n\(\geq\)-3
câu c
\(\Leftrightarrow\dfrac{2m-3}{2m+3}+\dfrac{2m+3}{2m-3}< 0\Leftrightarrow\dfrac{\left(2m-3\right)\left(2m-3\right)}{\left(2m+3\right)\left(2m-3\right)}+\dfrac{\left(2m+3\right)\left(2m+3\right)}{\left(2m-3\right)\left(2m+3\right)}< 0\)
\(\Leftrightarrow\dfrac{\left(2m-3\right)^2+\left(2m+3\right)^2}{\left(2m+3\right)\left(2m-3\right)}< 0\)
có
\(\left(2m-3\right)^2+\left(2m+3\right)^2>0\forall m\)
\(\Rightarrow\left(2m+3\right)\left(2m-3\right)< 0\Leftrightarrow\dfrac{-3}{2}< m< \dfrac{3}{2}\)
Cho biểu thức:
M=\(\left(2x+3\right)\left(2x-3\right)-2\left(x+5\right)^2-2\left(x-1\right)\left(x+2\right)\)
a) rút gọn M
b) tính giá trị của M tại \(x=-2\frac{1}{3}\)
c) Tìm x để M =0
a) M = ( 2x + 3)(2x - 3) - 2(x + 5)2 - 2(x - 1)(x + 2)
= 4x2 - 9 - 2(x2 + 10x + 25) - 2(x2 + x - 2)
= 4x2 - 9 - 2x2 - 20x - 50 - 2x2 - 2x + 4
= -22x - 55 = -11(2x + 5)
b) M = -11(2x + 5) = - 11(2.\(\frac{-7}{3}\)+ 5) = \(\frac{-11}{3}\)
b) M = -11(2x + 5) = 0
\(\Rightarrow\)2x + 5 = 0
\(\Rightarrow\)x = \(\frac{-5}{2}\)
Ta có: M = (2x+3)(2x-3) - 2(x+5)2 - 2(x-1)(x+2) \(=\left(2x\right)^2-3^2-2\left(x^2+10x+25\right)-\) \(2\left(x^2+x-2\right)\)
\(=4x^2-9-2x^2-20x-50-2x^2-2x+4\) =\(\left(4x^2-2x^2-2x^2\right)-\left(20x+2x\right)-\left(50+9-4\right)\) \(=-22x-55\)
b, Với x = \(-2\frac{1}{3}=\frac{-7}{3}\)
\(\Rightarrow M=-22.\frac{-7}{3}-55=\frac{154}{3}-55=\frac{-11}{3}\)
c, Để M = 0 => -22x - 55 = 0 \(\Rightarrow-22x=55\Rightarrow x=\frac{-55}{22}=\frac{-5}{2}\)
Vậy \(x=\frac{-5}{2}\)